Surface Integrals

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Describe surface S parameterized by

   \[\text{r}(u,v)=〈\text{cos}\ u,\text{sin}\ u,v〉,\text{-}\infty

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   To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of \({ℝ}^{2},\) we can choose any value for u and v and plot the corresponding point. If \(u=v=0,\) then \(\text{r}(0,0)=〈1,0,0〉,\) so point (1, 0, 0) is on S. Similarly, points \(\text{r}(\pi ,2)=(-1,0,2)\) and \(\text{r}(\frac{\pi }{2},4)=(0,1,4)\) are on S.

   Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize S, we visualize two families of curves that lie on S. In the first family of curves we hold u constant; in the second family of curves we hold v constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.

   First, suppose that u is a constant K. Then the curve traced out by the parameterization is \(〈\text{cos}\ K,\text{sin}\ K,v〉,\) which gives a vertical line that goes through point \((\text{cos}\ K,\text{sin}\ K,v)\) in the xy-plane.

   Now suppose that v is a constant K. Then the curve traced out by the parameterization is \(〈\text{cos}\ u,\text{sin}\ u,K〉,\) which gives a circle in plane \(z=K\) with radius 1 and center (0, 0, K).

   If u is held constant, then we get vertical lines; if v is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder \({x}^{2}+{y}^{2}=1\) ().

   Notice that if \(x=\text{cos}\ u\) and \(y=\text{sin}\ u,\) then \({x}^{2}+{y}^{2}=1,\) so points from S do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle \(〈\text{cos}\ u,\text{sin}\ u,k〉\) for some k, and therefore each point on the cylinder is contained in the parameterized surface ().

2. Describe the surface with parameterization \(\text{r}(u,v)=〈2\ \text{cos}\ u,2\ \text{sin}\ u,v〉,0\le u<2\pi ,\text{-}\infty

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   Cylinder \({x}^{2}+{y}^{2}=4\)

3. Describe surface S parameterized by

   \[\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,{u}^{2}〉,0\le u<\infty ,0\le v<2\pi .\]
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   Notice that if u is held constant, then the resulting curve is a circle of radius u in plane \(z={u}^{2}.\) Therefore, as u increases, the radius of the resulting circle increases. If v is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that

   \[\begin{array}{ll}{x}^{2}+{y}^{2} & ={(u\ \text{cos}\ v)}^{2}+{(u\ \text{sin}\ v)}^{2} \\ & ={u}^{2}{\text{cos}}^{2}v+{u}^{2}{\text{sin}}^{2}v \\ & ={u}^{2} \\ & =z.\end{array}\]

   Therefore, the surface is elliptic paraboloid \({x}^{2}+{y}^{2}=z\) ().

4. Describe the surface parameterized by \(\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,u〉,\text{-}\infty

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   Cone \({x}^{2}+{y}^{2}={z}^{2}\)

5. Give a parameterization of the cone \({x}^{2}+{y}^{2}={z}^{2}\) lying on or above the plane \(z=-2.\)

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   The horizontal cross-section of the cone at height \(z=u\) is circle \({x}^{2}+{y}^{2}={u}^{2}.\) Therefore, a point on the cone at height u has coordinates \((u\ \text{cos}\ v,u\ \text{sin}\ v,u)\) for angle v. Hence, a parameterization of the cone is \(\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,u〉.\) Since we are not interested in the entire cone, only the portion on or above plane \(z=-2,\) the parameter domain is given by \(-2\le u<\infty ,0\le v<2\pi\) ().

6. Give a parameterization for the portion of cone \({x}^{2}+{y}^{2}={z}^{2}\) lying in the first octant.

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   \(\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,u〉,\) \(0

7. Which of the figures in is smooth?

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   The surface in (a) can be parameterized by

   \[\text{r}(u,v)=〈(2+\text{cos}\ v)\text{cos}\ u,(2+\text{cos}\ v)\text{sin}\ u,\text{sin}\ v〉,0\le u<2\pi ,0\le v<2\pi\]

   (we can use technology to verify). Notice that vectors

   \[{\text{r}}_{u}=〈\text{-}(2+\text{cos}\ v)\text{sin}\ u,(2+\text{cos}\ v)\text{cos}\ u,0〉\ \text{and}\ {\text{r}}_{v}=〈\text{-}\text{sin}\ v\ \text{cos}\ u,\text{-}\text{sin}\ v\ \text{sin}\ u,\text{cos}\ v〉\]

   exist for any choice of u and v in the parameter domain, and

   \[\begin{array}{llllllll}{\text{r}}_{u}\ \times \ {\text{r}}_{v} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \text{-}(2+\text{cos}\ v)\text{sin}\ u & (2+\text{cos}\ v)\text{cos}\ u & 0 \\ \text{-}\text{sin}\ v\ \text{cos}\ u & \text{-}\text{sin}\ v\ \text{sin}\ u & \text{cos}\ v\end{array}| \\ & =[(2+\text{cos}\ v)\text{cos}\ u\ \text{cos}\ v]\text{i}+[(2+\text{cos}\ v)\text{sin}\ u\ \text{cos}\ v]\text{j} \\ & \ +[(2+\text{cos}\ v)\text{sin}\ v\ {\text{sin}}^{2}u+(2+\text{cos}\ v)\text{sin}\ v\ {\text{cos}}^{2}u]\text{k} \\ & =[(2+\text{cos}\ v)\text{cos}\ u\ \text{cos}\ v]\text{i}+[(2+\text{cos}\ v)\text{sin}\ u\ \text{cos}\ v]\text{j}+[(2+\text{cos}\ v)\text{sin}\ v]\text{k}.\end{array}\]

   The k component of this vector is zero only if \(v=0\) or \(v=\pi .\) If \(v=0\) or \(v=\pi ,\) then the only choices for u that make the j component zero are \(u=0\) or \(u=\pi .\) But, these choices of u do not make the i component zero. Therefore, \({\text{r}}_{u}\ \times \ {\text{r}}_{v}\) is not zero for any choice of u and v in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.

   In the pyramid in (b), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of five smooth faces, and thus this surface is piecewise smooth.

8. Is the surface parameterization \(\text{r}(u,v)=〈{u}^{2v},v+1,\text{sin}\ u〉,0\le u\le 2,0\le v\le 3\) smooth?

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   Yes

9. Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height h and radius r.

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   Before calculating the surface area of this cone using , we need a parameterization. We assume this cone is in \({ℝ}^{3}\) with its vertex at the origin (). To obtain a parameterization, let \(\alpha\) be the angle that is swept out by starting at the positive z-axis and ending at the cone, and let \(k=\text{tan}\ \alpha .\) For a height value v with \(0\le v\le h,\) the radius of the circle formed by intersecting the cone with plane \(z=v\) is \(kv.\) Therefore, a parameterization of this cone is

   \[\text{s}(u,v)=〈kv\ \text{cos}\ u,kv\ \text{sin}\ u,v〉,0\le u<2\pi ,0\le v\le h.\]

   The idea behind this parameterization is that for a fixed v value, the circle swept out by letting u vary is the circle at height v and radius kv. As v increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.

   With a parameterization in hand, we can calculate the surface area of the cone using . The tangent vectors are \({\text{t}}_{u}=〈\text{-}kv\ \text{sin}\ u,kv\ \text{cos}\ u,0〉\) and \({\text{t}}_{v}=〈k\ \text{cos}\ u,k\ \text{sin}\ u,1〉.\) Therefore,

   \[\begin{array}{llllllll}{\text{t}}_{u}\ \times \ {\text{t}}_{v} & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \text{-}kv\ \text{sin}\ u & kv\ \text{cos}\ u & 0 \\ k\ \text{cos}\ u & k\ \text{sin}\ u & 1\end{array}| \\ & =〈kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{-}{k}^{2}v\ {\text{sin}}^{2}u-{k}^{2}v\ {\text{cos}}^{2}u〉 \\ & =〈kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{-}{k}^{2}v〉.\end{array}\]

   The magnitude of this vector is

   \[\begin{array}{ll}‖〈kv\ \text{cos}\ u,kv\ \text{sin}\ u,\text{-}{k}^{2}v〉‖ & =\sqrt{{k}^{2}{v}^{2}{\text{cos}}^{2}u+{k}^{2}{v}^{2}{\text{sin}}^{2}u+{k}^{4}{v}^{2}} \\ & =\sqrt{{k}^{2}{v}^{2}+{k}^{4}{v}^{2}} \\ & =kv\sqrt{1+{k}^{2}}.\end{array}\]

   By , the surface area of the cone is

   \[\begin{array}{ll}{∬}_{D}‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖dA & ={\int }_{0}^{h}{\int }_{0}^{2\pi }kv\sqrt{1+{k}^{2}}dudv \\ & =2\pi k\sqrt{1+{k}^{2}}{\int }_{0}^{h}vdv \\ & =2\pi k\sqrt{1+{k}^{2}}{[\frac{{v}^{2}}{2}]}_{0}^{h} \\ & =\pi k{h}^{2}\sqrt{1+{k}^{2}}.\end{array}\]

   Since \(k=\text{tan}\ \alpha =r\text{/}h,\)

   \[\begin{array}{ll}\pi k{h}^{2}\sqrt{1+{k}^{2}} & =\pi \frac{r}{h}{h}^{2}\sqrt{1+\frac{{r}^{2}}{{h}^{2}}} \\ & =\pi rh\sqrt{1+\frac{{r}^{2}}{{h}^{2}}} \\ & =\pi r\sqrt{{h}^{2}+{h}^{2}(\frac{{r}^{2}}{{h}^{2}})} \\ & =\pi r\sqrt{{h}^{2}+{r}^{2}}.\end{array}\]

   Therefore, the lateral surface area of the cone is \(\pi r\sqrt{{h}^{2}+{r}^{2}}.\)

10. Find the surface area of the surface with parameterization \(\text{r}(u,v)=〈u+v,{u}^{2},2v〉,0\le u\le 3,0\le v\le 2.\)

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    \(\approx 43.02\)

11. Show that the surface area of the sphere \({x}^{2}+{y}^{2}+{z}^{2}={r}^{2}\) is \(4\pi {r}^{2}.\)

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    The sphere has parameterization

    \[〈r\ \text{cos}\ \theta \ \text{sin}\ ϕ,r\ \text{sin}\ \theta \ \text{sin}\ ϕ,r\ \text{cos}\ ϕ〉,0\le \theta <2\pi ,0\le ϕ\le \pi .\]

    The tangent vectors are

    \[{\text{t}}_{\theta }=〈\text{-}r\ \text{sin}\ \theta \ \text{sin}\ ϕ,r\ \text{cos}\ \theta \ \text{sin}\ ϕ,0〉\ \text{and}\ {\text{t}}_{ϕ}=〈r\ \text{cos}\ \theta \ \text{cos}\ ϕ,r\ \text{sin}\ \theta \ \text{cos}\ ϕ,\text{-}r\ \text{sin}\ ϕ〉.\]

    Therefore,

    \[\begin{array}{ll}{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta } & =〈{r}^{2}\text{cos}\ \theta \ {\text{sin}}^{2}ϕ,{r}^{2}\text{sin}\ \theta \ {\text{sin}}^{2}ϕ,{r}^{2}{\text{sin}}^{2}\theta \ \text{sin}\ ϕ\ \text{cos}\ ϕ+{r}^{2}{\text{cos}}^{2}\theta \ \text{sin}\ ϕ\ \text{cos}\ ϕ〉 \\ & =〈{r}^{2}\text{cos}\ \theta \ {\text{sin}}^{2}ϕ,{r}^{2}\text{sin}\ \theta \ {\text{sin}}^{2}ϕ,{r}^{2}\text{sin}\ ϕ\ \text{cos}\ ϕ〉.\end{array}\]

    Now,

    \[\begin{array}{ll}‖{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta }‖ & =\sqrt{{r}^{4}{\text{sin}}^{4}ϕ\ {\text{cos}}^{2}\theta +{r}^{4}{\text{sin}}^{4}ϕ\ {\text{sin}}^{2}\theta +{r}^{4}{\text{sin}}^{2}ϕ\ {\text{cos}}^{2}ϕ} \\ & =\sqrt{{r}^{4}{\text{sin}}^{4}ϕ+{r}^{4}{\text{sin}}^{2}ϕ\ {\text{cos}}^{2}ϕ} \\ & ={r}^{2}\sqrt{{\text{sin}}^{2}ϕ} \\ & ={r}^{2}\ \text{sin}\ ϕ.\end{array}\]

    Notice that \(\text{sin}\ ϕ\ge 0\) on the parameter domain because \(0\le ϕ<\pi ,\) and this justifies equation \(\sqrt{{\text{sin}}^{2}ϕ}=\text{sin}\ ϕ.\) The surface area of the sphere is

    \[{\int }_{0}^{2\pi }{\int }_{0}^{\pi }{r}^{2}\text{sin}\ ϕdϕd\theta ={r}^{2}{\int }_{0}^{2\pi }2d\theta =4\pi {r}^{2}.\]

    We have derived the familiar formula for the surface area of a sphere using surface integrals.

12. Show that the surface area of cylinder \({x}^{2}+{y}^{2}={r}^{2},0\le z\le h\) is \(2\pi rh.\) Notice that this cylinder does not include the top and bottom circles.

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    With the standard parameterization of a cylinder, shows that the surface area is \(2\pi rh.\)

13. Find the area of the surface of revolution obtained by rotating \(y={x}^{2},0\le x\le b\) about the x-axis ().

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    This surface has parameterization

    \[\text{r}(x,\theta )=〈x,{x}^{2}\text{cos}\ \theta ,{x}^{2}\text{sin}\ \theta 〉,0\le x\le b,0\le \theta <2\pi .\]

    The tangent vectors are \({\text{t}}_{x}=〈1,2x\ \text{cos}\ \theta ,2x\ \text{sin}\ \theta 〉\ \text{and}\ {\text{t}}_{\theta }=〈0,\text{-}{x}^{2}\text{sin}\ \theta ,{x}^{2}\text{cos}\ \theta 〉.\) Therefore,

    \[\begin{array}{ll}{\text{t}}_{x}\ \times \ {\text{t}}_{\theta } & =〈2{x}^{3}{\text{cos}}^{2}\theta +2{x}^{3}{\text{sin}}^{2}\theta ,\text{-}{x}^{2}\text{cos}\ \theta ,\text{-}{x}^{2}\text{sin}\ \theta 〉 \\ & =〈2{x}^{3},\text{-}{x}^{2}\text{cos}\ \theta ,\text{-}{x}^{2}\text{sin}\ \theta 〉\end{array}\]

    and

    \[\begin{array}{ll}||{\text{t}}_{x}\ \times \ {\text{t}}_{\theta }|| & =\sqrt{4{x}^{6}+{x}^{4}{\text{cos}}^{2}\theta +{x}^{4}{\text{sin}}^{2}\theta } \\ & =\sqrt{4{x}^{6}+{x}^{4}} \\ & ={x}^{2}\sqrt{4{x}^{2}+1}.\end{array}\]

    The area of the surface of revolution is

    \[\begin{array}{ll}{\int }_{0}^{b}{\int }_{0}^{\pi }{x}^{2}\sqrt{4{x}^{2}+1}d\theta dx & =2\pi {\int }_{0}^{b}{x}^{2}\sqrt{4{x}^{2}+1}dx \\ & =2\pi {[\frac{1}{64}(2\sqrt{4{x}^{2}+1}(8{x}^{3}+x){-\text{sinh}}^{-1}(2x))]}_{0}^{b} \\ & =2\pi [\frac{1}{64}(2\sqrt{4{b}^{2}+1}(8{b}^{3}+b){-\text{sinh}}^{-1}(2b))].\end{array}\]

14. Use to find the area of the surface of revolution obtained by rotating curve \(y=\text{sin}\ x,0\le x\le \pi\) about the x-axis.

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    \(2\pi (\sqrt{2}+{\text{sinh}}^{-1}(1))\)

15. Calculate surface integral \({∬}_{S}5dS,\) where \(S\) is the surface with parameterization \(\text{r}(u,v)=〈u,{u}^{2},v〉\) for \(0\le u\le 2\) and \(0\le v\le u.\)

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    Notice that this parameter domain D is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because does not place any restrictions on the shape of the parameter domain.

    To use to calculate the surface integral, we first find vector \({\text{t}}_{u}\) and \({\text{t}}_{v}.\) Note that \({\text{t}}_{u}=〈1,2u,0〉\) and \({\text{t}}_{v}=〈0,0,1〉.\) Therefore,

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ 1 & 2u & 0 \\ 0 & 0 & 1\end{array}|=〈2u,-1,0〉\]

    and

    \[‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖=\sqrt{1+4{u}^{2}}.\]

    By ,

    \[\begin{array}{ll}{∬}_{S}5dS & =5{∬}_{D}\sqrt{1+4{u}^{2}}dA \\ & =5{\int }_{0}^{2}{\int }_{0}^{u}\sqrt{1+4{u}^{2}}dvdu=5{\int }_{0}^{2}u\sqrt{1+4{u}^{2}}du \\ & =5{[\frac{{(1+4{u}^{2})}^{3\text{/}2}}{3}]}_{0}^{2}=\frac{5({17}^{3\text{/}2}-1)}{12}\approx 28.79.\end{array}\]

16. Calculate surface integral \({∬}_{S}(x+{y}^{2})dS,\) where S is cylinder \({x}^{2}+{y}^{2}=4,0\le z\le 3\) ().

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    To calculate the surface integral, we first need a parameterization of the cylinder. Following , a parameterization is

    \[\text{r}(u,v)=〈2\text{cos}\ u,2\text{sin}\ u,v〉,0\le u\le 2\pi ,0\le v\le 3.\]

    The tangent vectors are \({\text{t}}_{u}=〈-2\text{sin}\ u,2\text{cos}\ u,0〉\) and \({\text{t}}_{v}=〈0,0,1〉.\) Then,

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ -2\text{sin}\ u & 2\text{cos}\ u & 0 \\ 0 & 0 & 1\end{array}|=⟨2\text{cos}\ u,2\text{sin}\ u,0⟩\]

    and \(‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖=\sqrt{{4\text{cos}}^{2}u+{4\text{sin}}^{2}u}=2.\) By ,

    \[\begin{array}{l}\ \\ \\ \\ {∬}_{S}f(x,y,z)dS={∬}_{D}f(\text{r}(u,v))‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖\ dA \\ ={\int }_{0}^{3}{\int }_{0}^{2\pi }(2\text{cos}\ u+4{\text{sin}}^{2}u)2dudv \\ =2{\int }_{0}^{3}{[2\text{sin}\ u+2u-\sin (2u)]}_{0}^{2\pi }dv=2{\int }_{0}^{3}4\pi dv=24\pi .\end{array}\]

17. Calculate \({∬}_{S}({x}^{2}-z)dS,\) where S is the surface with parameterization \(\text{r}(u,v)=〈v,{u}^{2}+{v}^{2},1〉,0\le u\le 2,0\le v\le 3.\)

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    24

18. Calculate surface integral \({∬}_{S}f(x,y,z)dS,\) where \(f(x,y,z)={z}^{2}\) and S is the surface that consists of the piece of sphere \({x}^{2}+{y}^{2}+{z}^{2}=4\) that lies on or above plane \(z=1\) and the disk that is enclosed by intersection plane \(z=1\) and the given sphere ().

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    Notice that S is not smooth but is piecewise smooth; S can be written as the union of its base \({S}_{1}\) and its spherical top \({S}_{2},\) and both \({S}_{1}\) and \({S}_{2}\) are smooth. Therefore, to calculate \({∬}_{S}{z}^{2}dS,\) we write this integral as \({∬}_{{S}_{1}}{z}^{2}dS+{∬}_{{S}_{2}}{z}^{2}dS\) and we calculate integrals \({∬}_{{S}_{1}}{z}^{2}dS\) and \({∬}_{{S}_{2}}{z}^{2}dS.\)

    First, we calculate \({∬}_{{S}_{1}}{z}^{2}dS.\) To calculate this integral we need a parameterization of \({S}_{1}.\) This surface is a disk in plane \(z=1\) centered at \((0,0,1).\) To parameterize this disk, we need to know its radius. Since the disk is formed where plane \(z=1\) intersects sphere \({x}^{2}+{y}^{2}+{z}^{2}=4,\) we can substitute \(z=1\) into equation \({x}^{2}+{y}^{2}+{z}^{2}=4\text{:}\)

    \[{x}^{2}+{y}^{2}+1=4⇒{x}^{2}+{y}^{2}=3.\]

    Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \({S}_{1}\) is \(\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,1〉,0\le u\le \sqrt{3},0\le v\le 2\pi .\) The tangent vectors are \({\text{t}}_{u}=〈\text{cos}\ v,\text{sin}\ v,0〉\) and \({\text{t}}_{v}=〈\text{-}u\ \text{sin}\ v,u\text{co}sv,0〉,\) and thus

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ \text{cos}\ v & \text{sin}\ v & 0 \\ \text{-}u\ \text{sin}\ v & u\ \text{cos}\ v & 0\end{array}|=〈0,0,u\ {\text{cos}}^{2}v+u\ {\text{sin}}^{2}v〉=〈0,0,u〉.\]

    The magnitude of this vector is u. Therefore,

    \[\begin{array}{ll}{∬}_{{S}_{1}}{z}^{2}dS & ={\int }_{0}^{\sqrt{3}}{\int }_{0}^{2\pi }f(\text{r}(u,v))‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖\ dv\ du \\ & ={\int }_{0}^{\sqrt{3}}{\int }_{0}^{2\pi }u\ dv\ du \\ & =2\pi {\int }_{0}^{\sqrt{3}}udu \\ & =3\pi .\end{array}\]

    Now we calculate \({∬}_{{S}_{2}}dS.\) To calculate this integral, we need a parameterization of \({S}_{2}.\) The parameterization of full sphere \({x}^{2}+{y}^{2}+{z}^{2}=4\) is

    \[\text{r}(ϕ,\theta )=〈2\ \text{cos}\ \theta \ \text{sin}\ ϕ,2\ \text{sin}\ \theta \ \text{sin}\ ϕ,2\ \text{cos}\ ϕ〉,0\le \theta \le 2\pi ,0\le ϕ\le \pi .\]

    Since we are only taking the piece of the sphere on or above plane \(z=1,\) we have to restrict the domain of \(ϕ.\) To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in (the \(\sqrt{3}\) comes from the fact that the base of S is a disk with radius \(\sqrt{3}).\) Therefore, the tangent of \(ϕ\) is \(\sqrt{3},\) which implies that \(ϕ\) is \(\pi \text{/}3.\) We now have a parameterization of \({S}_{2}\text{:}\)

    \[\text{r}(ϕ,\theta )=〈2\ \text{cos}\ \theta \ \text{sin}\ ϕ,2\ \text{sin}\ \theta \ \text{sin}\ ϕ,2\ \text{cos}\ ϕ〉,0\le \theta \le 2\pi ,0\le ϕ\le \pi \text{/}3.\]

    The tangent vectors are

    \[{\text{t}}_{ϕ}=〈2\ \text{cos}\ \theta \ \text{cos}\ ϕ,2\ \text{sin}\ \theta \ \text{cos}\ ϕ,-2\ \text{sin}\ ϕ〉\ \text{and}\ {\text{t}}_{\theta }=〈-2\ \text{sin}\ \theta \ \text{sin}\ ϕ,u\ \text{cos}\ \theta \ \text{sin}\ ϕ,0〉,\]

    and thus

    \[\begin{array}{llllllll}{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta } & =|\begin{array}{lll}\text{i} & \text{j} & \text{k} \\ 2\ \text{cos}\ \theta \ \text{cos}\ ϕ & 2\ \text{sin}\ \theta \ \text{cos}\ ϕ & -2\ \text{sin}\ ϕ \\ -2\ \text{sin}\ \theta \ \text{sin}\ ϕ & 2\ \text{cos}\ \theta \ \text{sin}\ ϕ & 0\end{array}| \\ & =〈4\ \text{cos}\ \theta \ {\text{sin}}^{2}ϕ,4\ \text{sin}\ \theta \ {\text{sin}}^{2}ϕ,4\ {\text{cos}}^{2}\theta \ \text{cos}\ ϕ\ \text{sin}\ ϕ+4\ {\text{sin}}^{2}\theta \ \text{cos}\ ϕ\ \text{sin}\ ϕ〉 \\ & =〈4\ \text{cos}\ \theta \ {\text{sin}}^{2}ϕ,4\ \text{sin}\ \theta \ {\text{sin}}^{2}ϕ,4\ \text{cos}\ ϕ\ \text{sin}\ ϕ〉.\end{array}\]

    The magnitude of this vector is

    \[\begin{array}{ll}‖{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta }‖ & =\sqrt{16\ {\text{cos}}^{2}\theta \ {\text{sin}}^{4}ϕ+16\ {\text{sin}}^{2}\theta \ {\text{sin}}^{4}ϕ+16\ {\text{cos}}^{2}ϕ\ {\text{sin}}^{2}ϕ} \\ & =4\sqrt{{\text{sin}}^{4}ϕ+{\text{cos}}^{2}ϕ\ {\text{sin}}^{2}ϕ}.\end{array}\]

    Therefore,

    \[\begin{array}{ll}{∬}_{{S}_{2}}zdS & ={\int }_{0}^{\pi \text{/3}}{\int }_{0}^{2\pi }f(\text{r}(ϕ,\theta ))‖{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta }‖\ d\theta \ dϕ \\ & ={\int }_{0}^{\pi \text{/3}}{\int }_{0}^{2\pi }16\ {\text{cos}}^{2}ϕ\sqrt{{\text{sin}}^{4}ϕ+{\text{cos}}^{2}ϕ\ {\text{sin}}^{2}ϕ}d\theta \ dϕ \\ & =32\pi {\int }_{0}^{\pi \text{/3}}{\text{cos}}^{2}ϕ\sqrt{{\text{sin}}^{4}ϕ+{\text{cos}}^{2}ϕ\ {\text{sin}}^{2}ϕ}\ dϕ \\ & =32\pi {\int }_{0}^{\pi \text{/3}}{\text{cos}}^{2}ϕ\ \text{sin}\ ϕ\sqrt{{\text{sin}}^{2}ϕ+{\text{cos}}^{2}ϕ}\ dϕ \\ & =32\pi {\int }_{0}^{\pi \text{/3}}{\text{cos}}^{2}ϕ\ \text{sin}\ ϕ\ dϕ \\ & =32\pi {[-\frac{{\text{cos}}^{3}ϕ}{3}]}_{0}^{\pi \text{/}3}=32\pi [\frac{1}{3}-\frac{\sqrt{3}}{8}]=\frac{28\pi }{3}.\end{array}\]

    Since \({∬}_{S}{z}^{2}dS={∬}_{{S}_{1}}{z}^{2}dS+{∬}_{{S}_{2}}{z}^{2}dS=3\pi +\frac{28\pi }{3}=\frac{37\pi }{3}\)

19. Calculate surface integral \({∬}_{S}(x-y)dS,\) where S is cylinder \({x}^{2}+{y}^{2}=1,0\le z\le 2,\) including the circular top and bottom.

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    0

20. A flat sheet of metal has the shape of surface \(z=1+x+2y\) that lies above rectangle \(0\le x\le 4\) and \(0\le y\le 2.\) If the density of the sheet is given by \(\rho (x,y,z)={x}^{2}yz,\) what is the mass of the sheet?

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    Let S be the surface that describes the sheet. Then, the mass of the sheet is given by \(m={∬}_{S}{x}^{2}yzdS.\) To compute this surface integral, we first need a parameterization of S. Since S is given by the function \(f(x,y)=1+x+2y,\) a parameterization of S is \(\text{r}(x,y)=〈x,y,1+x+2y〉,0\le x\le 4,0\le y\le 2.\)

    The tangent vectors are \({\text{t}}_{x}=〈1,0,1〉\) and \({\text{t}}_{y}=〈1,0,2〉.\) Therefore, \({\text{t}}_{x}\ \times \ {\text{t}}_{y}=〈-1,-2,1〉\) and \(‖{\text{t}}_{x}\ \times \ {\text{t}}_{y}‖=\sqrt{6}.\) By ,

    \[\begin{array}{ll}m & ={∬}_{S}{x}^{2}y{z}^{}dS \\ & =\sqrt{6}{\int }_{0}^{4}{\int }_{0}^{2}{x}^{2}y(1+x+2y)dydx \\ & =\sqrt{6}{\int }_{0}^{4}\frac{22{x}^{2}}{3}+2{x}^{3}dx \\ & =\frac{2560\sqrt{6}}{9} \\ & \approx 696.74.\end{array}\]

21. A piece of metal has a shape that is modeled by paraboloid \(z={x}^{2}+{y}^{2},0\le z\le 4,\) and the density of the metal is given by \(\rho (x,y,z)=z+1.\) Find the mass of the piece of metal.

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    \(38.401\pi \approx 120.640\)

22. Give an orientation of cylinder \({x}^{2}+{y}^{2}={r}^{2},0\le z\le h.\)

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    This surface has parameterization

    \[\text{r}(u,v)=〈r\ \text{cos}\ u,r\ \text{sin}\ u,v〉,0\le u<2\pi ,0\le v\le h.\]

    The tangent vectors are \({\text{t}}_{u}=〈\text{-}r\ \text{sin}\ u,r\ \text{cos}\ u,0〉\) and \({\text{t}}_{v}=〈0,0,1〉.\) To get an orientation of the surface, we compute the unit normal vector

    \[\text{N}=\frac{{\text{t}}_{u}\ \times \ {\text{t}}_{v}}{‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖}.\]

    In this case, \({\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈r\ \text{cos}\ u,r\ \text{sin}\ u,0〉\) and therefore

    \[‖{\text{t}}_{u}\ \times \ {\text{t}}_{v}‖=\sqrt{{r}^{2}{\text{cos}}^{2}u+{r}^{2}{\text{sin}}^{2}u}=r.\]

    An orientation of the cylinder is

    \[\text{N}(u,v)=\frac{〈r\ \text{cos}\ u,r\ \text{sin}\ u,0〉}{r}=〈\text{cos}\ u,\text{sin}\ u,0〉.\]

    Notice that all vectors are parallel to the xy-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder ().

23. Give the “upward” orientation of the graph of \(f(x,y)=xy.\)

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    \(\text{N}(x,y)=〈\frac{\text{-}y}{\sqrt{1+{x}^{2}+{y}^{2}}},\frac{\text{-}x}{\sqrt{1+{x}^{2}+{y}^{2}}},\frac{1}{\sqrt{1+{x}^{2}+{y}^{2}}}〉\)

24. Calculate the surface integral \({∬}_{S}\text{F}\cdot \text{N}dS,\) where \(\text{F}=〈\text{-}y,x,0〉\) and \(S\) is the surface with parameterization \(\text{r}(u,v)=〈u,{v}^{2}-u,u+v〉,0\le u<3,0\le v\le 4.\)

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    The tangent vectors are \({\text{t}}_{u}=〈1,-1,1〉\) and \({\text{t}}_{v}=〈0,2v,1〉.\) Therefore,

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈-1-2v,-1,2v〉.\]

    By ,

    \[\begin{array}{ll}{∬}_{S}\text{F}\cdot d\text{S} & ={\int }_{0}^{4}{\int }_{0}^{3}\text{F}(\text{r}(u,v))\cdot ({\text{t}}_{u}\ \times \ {\text{t}}_{v})dudv \\ & ={\int }_{0}^{4}{\int }_{0}^{3}〈u-{v}^{2},u,0〉\cdot 〈-1-2v,-1,2v〉\ dudv \\ & ={\int }_{0}^{4}{\int }_{0}^{3}[(u-{v}^{2})(-1-2v)-u]dudv \\ & ={\int }_{0}^{4}{\int }_{0}^{3}(2{v}^{3}+{v}^{2}-2uv-2u)dudv \\ & ={{\int }_{0}^{4}[2{v}^{3}u+{v}^{2}u-v{u}^{2}-{u}^{2}]}_{0}^{3}dv \\ & ={\int }_{0}^{4}(6{v}^{3}+3{v}^{2}-9v-9)dv \\ & ={[\frac{3{v}^{4}}{2}+{v}^{3}-\frac{9{v}^{2}}{2}-9v]}_{0}^{4} \\ & =340.\end{array}\]

    Therefore, the flux of F across S is 340.

25. Calculate surface integral \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}=〈0,\text{-}z,y〉\) and S is the portion of the unit sphere in the first octant with outward orientation.

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    0

26. Let \(\text{v}(x,y,z)=〈2x,2y,z〉\) represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m³. Let S be hemisphere \({x}^{2}+{y}^{2}+{z}^{2}=9\) with \(z\ge 0\) such that S is oriented outward. Find the mass flow rate of the fluid across S.

    Avslöja svaret

    A parameterization of the surface is

    \[\text{r}(ϕ,\theta )=〈3\ \text{cos}\ \theta \ \text{sin}\ ϕ,3\ \text{sin}\ \theta \ \text{sin}\ ϕ,3\ \text{cos}\ ϕ〉,0\le \theta \le 2\pi ,0\le ϕ\le \pi \text{/}2.\]

    As in , the tangent vectors are

    \[{\text{t}}_{\theta }=〈-3\ \text{sin}\ \theta \ \text{sin}\ ϕ,3\ \text{cos}\ \theta \ \text{sin}\ ϕ,0〉\ \text{and}\ {\text{t}}_{ϕ}=〈3\ \text{cos}\ \theta \ \text{cos}\ ϕ,3\ \text{sin}\ \theta \ \text{cos}\ ϕ,-3\ \text{sin}\ ϕ〉,\]

    and their cross product is

    \[{\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta }=〈9\ \text{cos}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ ϕ\ \text{cos}\ ϕ〉.\]

    Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation \(\text{N}=〈9\ \text{cos}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ ϕ\ \text{cos}\ ϕ〉\) for the sphere.

    By ,

    \[\begin{array}{ll}{∬}_{S}\rho \text{v}\cdot d\text{S} & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2}\text{v}(\text{r}(ϕ,\theta ))\cdot ({\text{t}}_{ϕ}\ \times \ {\text{t}}_{\theta })dϕd\theta \\ & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2} \\ & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2}\begin{array}{l}〈6\ \text{cos}\ \theta \ \text{sin}\ ϕ,6\ \text{sin}\ \theta \ \text{sin}\ ϕ,3\ \text{cos}\ ϕ〉 \\ \cdot 〈9\ \text{cos}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ \theta \ {\text{sin}}^{2}ϕ,9\ \text{sin}\ ϕ\ \text{cos}\ ϕ〉dϕd\theta \end{array} \\ & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2}54\ {\text{sin}}^{3}ϕ+27\ {\text{cos}}^{2}ϕ\ \text{sin}\ ϕdϕd\theta \\ & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2}54(1-{\text{cos}}^{2}ϕ)\text{sin}\ ϕ+27\ {\text{cos}}^{2}ϕ\ \text{sin}\ ϕdϕd\theta \\ & =80{\int }_{0}^{2\pi }{\int }_{0}^{\pi \text{/}2}54\ \text{sin}\ ϕ-27\ {\text{cos}}^{2}ϕ\ \text{sin}\ ϕdϕd\theta \\ & =80{\int }_{0}^{2\pi }{[-54\ \text{cos}\ ϕ+9\ {\text{cos}}^{3}ϕ]}_{ϕ=0}^{ϕ=2\pi }d\theta \\ & =80{\int }_{0}^{2\pi }45d\theta =7200\pi .\end{array}\]

    Therefore, the mass flow rate is \(7200\pi \ \text{kg}\text{/}\text{sec}\text{/}{\text{m}}^{2}.\)

27. Let \(\text{v}(x,y,z)=〈{x}^{2}+{y}^{2},z,4y〉\) m/sec represent a velocity field of a fluid with constant density 100 kg/m³. Let S be the half-cylinder \(\text{r}(u,v)=〈\text{cos}\ u,\text{sin}\ u,v〉,0\le u\le \pi ,0\le v\le 2\) oriented outward. Calculate the mass flux of the fluid across S.

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    400 kg/sec/m

28. A cast-iron solid cylinder is given by inequalities \({x}^{2}+{y}^{2}\le 1,\) \(1\le z\le 4.\) The temperature at point \((x,y,z)\) in a region containing the cylinder is \(T(x,y,z)=({x}^{2}+{y}^{2})z.\) Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

    Avslöja svaret

    Let S denote the boundary of the object. To find the heat flow, we need to calculate flux integral \({∬}_{S}\text{-}k∇T\cdot d\text{S}.\) Notice that S is not a smooth surface but is piecewise smooth, since S is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of S. Before calculating any integrals, note that the gradient of the temperature is \(∇T=〈2xz,2yz,{x}^{2}+{y}^{2}〉.\)

    First we consider the circular bottom of the object, which we denote \({S}_{1}.\) We can see that \({S}_{1}\) is a circle of radius 1 centered at point \((0,0,1),\) sitting in plane \(z=1.\) This surface has parameterization \(\text{r}(u,v)=〈v\ \text{cos}\ u,v\ \text{sin}\ u,1〉,0\le u<2\pi ,0\le v\le 1.\) Therefore,

    \[{\text{t}}_{u}=〈\text{-}v\ \text{sin}\ u,v\ \text{cos}\ u,0〉\ \text{and}\ {\text{t}}_{v}=〈\text{cos}\ u,v\ \text{sin}\ u,0〉,\]

    and

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈0,0,\text{-}v\ {\text{sin}}^{2}u-v\ {\text{cos}}^{2}u〉=〈0,0,\text{-}v〉.\]

    Since the surface is oriented outward and \({S}_{1}\) is the bottom of the object, it makes sense that this vector points downward. By , the heat flow across \({S}_{1}\) is

    \[\begin{array}{ll}{∬}_{{S}_{1}}\text{-}k∇T\cdot d\text{S} & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}∇T(u,v)\cdot ({\text{t}}_{u}\ \times \ {\text{t}}_{v})dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}〈2v\ \text{cos}\ u,2v\ \text{sin}\ u,{v}^{2}{\text{cos}}^{2}u+{v}^{2}{\text{sin}}^{2}u〉\cdot 〈0,0,\text{-}v〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}〈2v\ \text{cos}\ u,2v\ \text{sin}\ u,{v}^{2}〉\cdot 〈0,0,\text{-}v〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}\text{-}{v}^{3}dvdu=-55{\int }_{0}^{2\pi }-\frac{1}{4}du=\frac{55\pi }{2}.\end{array}\]

    Now let’s consider the circular top of the object, which we denote \({S}_{2}.\) We see that \({S}_{2}\) is a circle of radius 1 centered at point \((0,0,4),\) sitting in plane \(z=4.\) This surface has parameterization \(\text{r}(u,v)=〈v\ \text{cos}\ u,v\ \text{sin}\ u,4〉,0\le u<2\pi ,0\le v\le 1.\) Therefore,

    \[{\text{t}}_{u}=〈\text{-}v\ \text{sin}\ u,v\ \text{cos}\ u,0〉\ \text{and}\ {\text{t}}_{v}=〈\text{cos}\ u,v\ \text{sin}\ u,0〉,\]

    and

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈0,0,\text{-}v\ {\text{sin}}^{2}u-v\ {\text{cos}}^{2}u〉=〈0,0,\text{-}v〉.\]

    Since the surface is oriented outward and \({S}_{1}\) is the top of the object, we instead take vector \({\text{t}}_{v}\ \times \ {\text{t}}_{u}=〈0,0,v〉.\) By , the heat flow across \({S}_{1}\) is

    \[\begin{array}{ll}\int {\int }_{{S}_{2}}\text{-}k∇T\cdot d\text{S} & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}∇T(u,v)\cdot ({\text{t}}_{v}\ \times \ {\text{t}}_{u})dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}〈8v\ \text{cos}\ u,8v\ \text{sin}\ u,{v}^{2}{\text{cos}}^{2}u+{v}^{2}{\text{sin}}^{2}u〉\cdot 〈0,0,v〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}〈8v\ \text{cos}\ u,8v\ \text{sin}\ u,{v}^{2}〉\cdot 〈0,0,v〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}{v}^{3}dvdu=-\frac{55\pi }{2}.\end{array}\]

    Last, let’s consider the cylindrical side of the object. This surface has parameterization \(\text{r}(u,v)=〈\text{cos}\ u,\text{sin}\ u,v〉,0\le u<2\pi ,1\le v\le 4.\) By , we know that \({\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈\text{cos}\ u,\text{sin}\ u,0〉.\) By ,

    \[\begin{array}{ll}{∬}_{{S}_{3}}\text{-}k∇T\cdot d\text{S} & =-55{\int }_{0}^{2\pi }{\int }_{1}^{4}∇T(u,v)\cdot ({\text{t}}_{v}\ \times \ {\text{t}}_{u})dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{1}^{4}〈2v\ \text{cos}\ u,2v\ \text{sin}\ u,{\text{cos}}^{2}u+{\text{sin}}^{2}u〉\cdot 〈\text{cos}\ u,\text{sin}\ u,0〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}〈2v\ \text{cos}\ u,2v\ \text{sin}\ u,1〉\cdot 〈\text{cos}\ u,\text{sin}\ u,0〉dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}(2v\ {\text{cos}}^{2}u+2v\ {\text{sin}}^{2}u)dvdu \\ & =-55{\int }_{0}^{2\pi }{\int }_{0}^{1}2vdvdu=-55{\int }_{0}^{2\pi }du=-110\pi .\end{array}\]

    Therefore, the rate of heat flow across S is \(\frac{55\pi }{2}-\frac{55\pi }{2}-110\pi =-110\pi .\)

29. A cast-iron solid ball is given by inequality \({x}^{2}+{y}^{2}+{z}^{2}\le 1.\) The temperature at a point in a region containing the ball is \(T(x,y,z)=\frac{1}{3}({x}^{2}+{y}^{2}+{z}^{2}).\) Find the heat flow across the boundary of the solid if this boundary is oriented outward.

    Avslöja svaret

    \(-\frac{440\pi }{3}\)

30. If surface S is given by \(\{(x,y,z):0\le x\le 1,0\le y\le 1,z=10\},\) then \({∬}_{S}f(x,y,z)dS={\int }_{0}^{1}{\int }_{0}^{1}f(x,y,10)dxdy.\)

    Avslöja svaret

    True

31. If surface S is given by \(\{(x,y,z):0\le x\le 1,0\le y\le 1,z=x\},\) then \({∬}_{S}f(x,y,z)dS={\int }_{0}^{1}{\int }_{0}^{1}f(x,y,x)dxdy.\)

32. Surface \(\text{r}=〈v\ \text{cos}\ u,v\ \text{sin}\ u,{v}^{2}〉,\ \text{for}\ 0\le u\le \pi ,0\le v\le 2,\) is the same as surface \(\text{r}=〈\sqrt{v}\ \text{cos}\ 2u,\sqrt{v}\ \text{sin}\ 2u,v〉,\) for \(0\le u\le \frac{\pi }{2},0\le v\le 4.\)

    Avslöja svaret

    True

33. Given the standard parameterization of a sphere, normal vectors \({\text{t}}_{u}^{}\ \times \ {\text{t}}_{v}\) are outward normal vectors.

34. Plane \(3x-2y+z=2\)

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    \(\text{r}(u,v)=〈u,v,2-3u+2v〉\) for \(\text{-}\infty \le u<\infty\) and \(\text{-}\infty \le v<\infty .\)

35. Paraboloid \(z={x}^{2}+{y}^{2},\) for \(0\le z\le 9.\)

36. Plane \(2x-4y+3z=16\)

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    \(\text{r}(u,v)=〈u,v,\frac{1}{3}(16-2u+4v)〉\) for \(|u|<\infty\) and \(|v|<\infty .\)

37. The frustum of cone \({z}^{2}={x}^{2}+{y}^{2},\ \text{for}\ 2\le z\le 8\)

38. The portion of cylinder \({x}^{2}+{y}^{2}=9\) in the first octant, for \(0\le z\le 3\)

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    \(\text{r}(u,v)=〈3\ \text{cos}\ u,3\ \text{sin}\ u,v〉\) for \(0\le u\le \frac{\pi }{2},0\le v\le 3\)

39. A cone with base radius r and height h, where r and h are positive constants

40. [T] Half cylinder \(\{(r,\theta ,z):r=4,0\le \theta \le \pi ,0\le z\le 7\}\)

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    \(A=28\pi =87.9646\)

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Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mer information Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems