Change of Variables in Multiple 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. Suppose a transformation \(T\) is defined as \(T(r,\theta )=(x,y)\) where \(x=r\ \text{cos}\ \theta ,y=r\ \text{sin}\ \theta .\) Find the image of the polar rectangle \(G=\{(r,\theta )|0

   Jiżvelaw it-tweġiba

   Since \(r\) varies from 0 to 1 in the \(r\theta \text{-plane},\) we have a circular disc of radius 0 to 1 in the \(xy\text{-plane}\text{.}\) Because \(\theta\) varies from 0 to \(\pi \text{/2}\) in the \(r\theta \text{-plane},\) we end up getting a quarter circle of radius \(1\) in the first quadrant of the \(xy\text{-plane}\) (). Hence \(R\) is a quarter circle bounded by \({x}^{2}+{y}^{2}=1\) in the first quadrant.

   In order to show that \(T\) is a one-to-one transformation, assume \(T({r}_{1},{\theta }_{1})=T({r}_{2},{\theta }_{2})\) and show as a consequence that \(({r}_{1},{\theta }_{1})=({r}_{2},{\theta }_{2}).\) In this case, we have

   \[\begin{array}{lll}T({r}_{1},{\theta }_{1}) & = & T({r}_{2},{\theta }_{2}), \\ ({x}_{1},{y}_{1}) & = & ({x}_{2},{y}_{2}), \\ ({r}_{1}\text{cos}\ {\theta }_{1},{r}_{1}\text{sin}\ {\theta }_{1}) & = & ({r}_{2}\text{cos}\ {\theta }_{2},{r}_{2}\text{sin}\ {\theta }_{2}), \\ {r}_{1}\text{cos}\ {\theta }_{1} & = & {r}_{2}\text{cos}\ {\theta }_{2} \\ {r}_{1}\text{sin}\ {\theta }_{1} & = & {r}_{2}\text{sin}\ {\theta }_{2}\end{array}\]

   Dividing, we obtain

   \[\begin{array}{lll}\frac{{r}_{1}\text{cos}\ {\theta }_{1}}{{r}_{1}\text{sin}\ {\theta }_{1}} & = & \frac{{r}_{2}\text{cos}\ {\theta }_{2}}{{r}_{2}\text{sin}\ {\theta }_{2}} \\ \frac{\text{cos}\ {\theta }_{1}}{\text{sin}\ {\theta }_{1}} & = & \frac{\text{cos}\ {\theta }_{2}}{\text{sin}\ {\theta }_{2}} \\ \text{cot}\ {\theta }_{1} & = & \text{cot}\ {\theta }_{2} \\ {\theta }_{1} & = & {\theta }_{2}\end{array}\]

   since the cotangent function is one-one function in the interval \(0\le \theta \le \pi \text{/}2.\) Also, since \(0

   To find \({T}^{-1}(x,y)\) solve for \(r,\theta\) in terms of \(x,y.\) We already know that \({r}^{2}={x}^{2}+{y}^{2}\) and \(\text{tan}\ \theta =\frac{y}{x}.\) Thus \({T}^{-1}(x,y)=(r,\theta )\) is defined as \(r=\sqrt{{x}^{2}+{y}^{2}}\) and \(\theta ={\text{tan}}^{-1}(\frac{y}{x}).\)

2. Let the transformation \(T\) be defined by \(T(u,v)=(x,y)\) where \(x={u}^{2}-{v}^{2}\) and \(y=uv.\) Find the image of the triangle in the \(uv\text{-plane}\) with vertices \((0,0),(0,1),\) and \((1,1).\)

   Jiżvelaw it-tweġiba

   The triangle and its image are shown in . To understand how the sides of the triangle transform, call the side that joins \((0,0)\) and \((0,1)\) side \(A,\) the side that joins \((0,0)\) and \((1,1)\) side \(B,\) and the side that joins \((1,1)\) and \((0,1)\) side \(C.\)

   For the side \(A\text{:}\ u=0,0\le v\le 1\) transforms to \(x=\text{-}{v}^{2},y=0\) so this is the side \(A'\) that joins \((-1,0)\) and \((0,0).\)

   For the side \(B\text{:}\ u=v,0\le u\le 1\) transforms to \(x=0,y={u}^{2}\) so this is the side \({B}^{'}\) that joins \((0,0)\) and \((0,1).\)

   For the side \(C\text{:}\ 0\le u\le 1,v=1\) transforms to \(x={u}^{2}-1,y=u\) (hence \(x={y}^{2}-1)\) so this is the side \({C}^{'}\) that makes the upper half of the parabolic arc joining \((-1,0)\) and \((0,1).\)

   All the points in the entire region of the triangle in the \(uv\text{-plane}\) are mapped inside the parabolic region in the \(xy\text{-plane}\text{.}\)

3. Let a transformation \(T\) be defined as \(T(u,v)=(x,y)\) where \(x=u+v,y=3v.\) Find the image of the rectangle \(G=\{(u,v)\text{:}\ 0\le u\le 1,0\le v\le 2\}\) from the \(uv\text{-plane}\) after the transformation into a region \(R\) in the \(xy\text{-plane}\text{.}\) Show that \(T\) is a one-to-one transformation and find \({T}^{-1}(x,y).\)

   Jiżvelaw it-tweġiba

   \({T}^{-1}(x,y)=(u,v)\) where \(u=\frac{3x-y}{3}\) and \(v=\frac{y}{3}\)

4. Find the Jacobian of the transformation given in .

   Jiżvelaw it-tweġiba

   The transformation in the example is \(T(r,\theta )=(r\ \text{cos}\ \theta ,r\ \text{sin}\ \theta )\) where \(x=r\ \text{cos}\ \theta\) and \(y=r\ \text{sin}\ \theta .\) Thus the Jacobian is

   \[\begin{array}{llllllllll}J(r,\theta ) & =\frac{∂(x,y)}{∂(r,\theta )}=|\begin{array}{lll}\frac{∂x}{∂r} & & \frac{∂x}{∂\theta } \\ \frac{∂y}{∂r} & & \frac{∂y}{∂\theta }\end{array}|=|\begin{array}{lll}\text{cos}\ \theta & & -r\ \text{sin}\ \theta \\ \text{sin}\ \theta & & r\ \text{cos}\ \theta \end{array}| \\ & =r\ {\text{cos}}^{2}\theta +r\ {\text{sin}}^{2}\theta =r({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )=r.\end{array}\]

5. Find the Jacobian of the transformation given in .

   Jiżvelaw it-tweġiba

   The transformation in the example is \(T(u,v)=({u}^{2}-{v}^{2},uv)\) where \(x={u}^{2}-{v}^{2}\) and \(y=uv.\) Thus the Jacobian is

   \[J(u,v)=\frac{∂(x,y)}{∂(u,v)}=|\begin{array}{lll}\frac{∂x}{∂u} & & \frac{∂x}{∂v} \\ \frac{∂y}{∂u} & & \frac{∂y}{∂v}\end{array}|=|\begin{array}{lll}2u & & v \\ -2v & & u\end{array}|=2{u}^{2}+2{v}^{2}.\]

6. Find the Jacobian of the transformation: \(T(u,v)=(u+v,2v).\)

   Jiżvelaw it-tweġiba

   \(J(u,v)=\frac{∂(x,y)}{∂(u,v)}=|\begin{array}{lll}\frac{∂x}{∂u} & & \frac{∂x}{∂v} \\ \frac{∂y}{∂u} & & \frac{∂y}{∂v}\end{array}|=|\begin{array}{lll}1 & & 1 \\ 0 & & 2\end{array}|=2\)

7. Consider the integral

   \[\int _{0}^{2}\ \int _{0}^{\sqrt{2x-{x}^{2}}}\sqrt{{x}^{2}+{y}^{2}}\ dy\ dx.\]

   Use the change of variables \(x=r\ \text{cos}\ \theta\) and \(y=r\ \text{sin}\ \theta ,\) and find the resulting integral.

   Jiżvelaw it-tweġiba

   First we need to find the region of integration. This region is bounded below by \(y=0\) and above by \(y=\sqrt{2x-{x}^{2}}\) (see the following figure).

   Squaring and collecting terms, we find that the region is the upper half of the circle \({x}^{2}+{y}^{2}-2x=0,\) that is, \({y}^{2}+{(x-1)}^{2}=1.\) In polar coordinates, the circle is \(r=2\ \text{cos}\ \theta\) so the region of integration in polar coordinates is bounded by \(0\le r\le 2\text{cos}\ \theta\) and \(0\le \theta \le \frac{\pi }{2}.\)

   The Jacobian is \(J(r,\theta )=r,\) as shown in . Since \(r\ge 0,\) we have \(|J(r,\theta )|=r.\)

   The integrand \(\sqrt{{x}^{2}+{y}^{2}}\) changes to \(r\) in polar coordinates, so the double iterated integral is

   \[\int _{0}^{2}\ \int _{0}^{\sqrt{2x-{x}^{2}}}\sqrt{{x}^{2}+{y}^{2}}dy\ dx=\int _{0}^{\pi \text{/}2}\ \int _{0}^{2\ \text{cos}\ \theta }r|J(r,\theta )|dr\ d\theta =\int _{0}^{\pi \text{/}2}\ \int _{0}^{2\ \text{cos}\ \theta }{r}^{2}dr\ d\theta .\]

8. Considering the integral \(\int _{0}^{1}\ \int _{0}^{\sqrt{1-{x}^{2}}}({x}^{2}+{y}^{2})dy\ dx,\) use the change of variables \(x=r\ \text{cos}\ \theta\) and \(y=r\ \text{sin}\ \theta ,\) and find the resulting integral.

   Jiżvelaw it-tweġiba

   \(\int _{0}^{\pi \text{/}2}\ \int _{0}^{1}{r}^{3}dr\ d\theta\)

9. Consider the integral \(\underset{R}{∬}(x-y)dy\ dx,\) where \(R\) is the parallelogram joining the points \((1,2),\) \((3,4),(4,3),\) and \((6,5)\) (). Make appropriate changes of variables, and write the resulting integral.

   Jiżvelaw it-tweġiba

   First, we need to understand the region over which we are to integrate. The sides of the parallelogram are \(x-y+1=0,x-y-1=0,\) \(x-3y+5=0,\text{and}\ x-3y+9=0\) (). Another way to look at them is \(x-y=-1,x-y=1,\) \(x-3y=-5,\) and \(x-3y=-9.\)

   Clearly the parallelogram is bounded by the lines \(y=x+1,y=x-1,y=\frac{1}{3}(x+5),\) and \(y=\frac{1}{3}(x+9).\)

   Notice that if we were to make \(u=x-y\) and \(v=x-3y,\) then the limits on the integral would be \(-1\le u\le 1\) and \(-9\le v\le -5.\)

   To solve for \(x\) and \(y,\) we multiply the first equation by \(3\) and subtract the second equation, \(3u-v=(3x-3y)-(x-3y)=2x.\) Then we have \(x=\frac{3u-v}{2}.\) Moreover, if we simply subtract the second equation from the first, we get \(u-v=(x-y)-(x-3y)=2y\) and \(y=\frac{u-v}{2}.\)

   Thus, we can choose the transformation

   \[T(u,v)=(\frac{3u-v}{2},\frac{u-v}{2})\]

   and compute the Jacobian \(J(u,v).\) We have

   \[J(u,v)=\frac{∂(x,y)}{∂(u,v)}=|\begin{array}{lll}\frac{∂x}{∂u} & & \frac{∂x}{∂v} \\ \frac{∂y}{∂u} & & \frac{∂y}{∂v}\end{array}|=|\begin{array}{lll}3\text{/}2 & & -1\text{/}2 \\ 1\text{/}2 & & -1\text{/}2\end{array}|=-\frac{3}{4}+\frac{1}{4}=-\frac{1}{2}.\]

   Therefore, \(|J(u,v)|=\frac{1}{2}.\) Also, the original integrand becomes

   \[x-y=\frac{1}{2}[3u-v-u+v]=\frac{1}{2}[3u-u]=\frac{1}{2}[2u]=u.\]

   Therefore, by the use of the transformation \(T,\) the integral changes to

   \[\underset{R}{∬}(x-y)dy\ dx=\int _{-9}^{-5}\ \int _{-1}^{1}J(u,v)u\ du\ dv=\int _{-9}^{-5}\ \int _{-1}^{1}(\frac{1}{2})u\ du\ dv,\]

   which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.

10. Make appropriate changes of variables in the integral \(\underset{R}{∬}\frac{4}{{(x-y)}^{2}}dy\ dx,\) where \(R\) is the trapezoid bounded by the lines \(x-y=2,x-y=4,x=0,\text{and}\ y=0.\) Write the resulting integral.

    Jiżvelaw it-tweġiba

    \(x=\frac{1}{2}(v+u)\) and \(y=\frac{1}{2}(v-u)\) and \(\int _{2}^{4}\ \int _{-u}^{u}\frac{4}{{u}^{2}}(\frac{1}{2})dv\ du.\)

11. Using the change of variables \(u=x-y\) and \(v=x+y,\) evaluate the integral

    \[\underset{R}{∬}(x-y){e}^{{x}^{2}-{y}^{2}}dA,\]

    where \(R\) is the region bounded by the lines \(x+y=1\) and \(x+y=3\) and the curves \({x}^{2}-{y}^{2}=-1\) and \({x}^{2}-{y}^{2}=1\) (see the first region in ).

    Jiżvelaw it-tweġiba

    As before, first find the region \(R\) and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made ().

    Given \(u=x-y\) and \(v=x+y,\) we have \(x=\frac{u+v}{2}\) and \(y=\frac{v-u}{2}\) and hence the transformation to use is \(T(u,v)=(\frac{u+v}{2},\frac{v-u}{2}).\) The lines \(x+y=1\) and \(x+y=3\) become \(v=1\) and \(v=3,\) respectively. The curves \({x}^{2}-{y}^{2}=1\) and \({x}^{2}-{y}^{2}=-1\) become \(uv=1\) and \(uv=-1,\) respectively.

    Thus we can describe the region \(S\) (see the second region ) as

    \[S=\{(u,v)|1\le v\le 3,\frac{-1}{v}\le u\le \frac{1}{v}\}.\]

    The Jacobian for this transformation is

    \[J(u,v)=\frac{∂(x,y)}{∂(u,v)}=|\begin{array}{lll}\frac{∂x}{∂u} & & \frac{∂x}{∂v} \\ \frac{∂y}{∂u} & & \frac{∂y}{∂v}\end{array}|=|\begin{array}{lll}1\text{/}2 & & -1\text{/}2 \\ 1\text{/}2 & & 1\text{/}2\end{array}|=\frac{1}{2}.\]

    Therefore, by using the transformation \(T,\) the integral changes to

    \[\underset{R}{∬}(x-y){e}^{{x}^{2}-{y}^{2}}dA=\frac{1}{2}\int _{1}^{3}\ \int _{-1\text{/}v}^{1\text{/}v}u{e}^{uv}du\ dv.\]

    Doing the evaluation, we have

    \[\frac{1}{2}\int _{1}^{3}\ \int _{-1\text{/}v}^{1\text{/}v}u{e}^{uv}du\ dv=\frac{2}{3e}\approx 0.245.\]

12. Using the substitutions \(x=v\) and \(y=\sqrt{u+v},\) evaluate the integral \(\underset{R}{∬}y\ \text{sin}({y}^{2}-x)dA\) where \(R\) is the region bounded by the lines \(y=\sqrt{x},x=2,\text{and}\ y=0.\)

    Jiżvelaw it-tweġiba

    \(\frac{1}{2}(\text{sin}\ 2-2)\)

13. Derive the formula in triple integrals for

    1. cylindrical and
    2. spherical coordinates.
    Jiżvelaw it-tweġiba

    1. For cylindrical coordinates, the transformation is \(T(r,\theta ,z)=(x,y,z)\) from the Cartesian \(r\theta z\text{-plane}\) to the Cartesian \(xyz\text{-plane}\) (). Here \(x=r\ \text{cos}\ \theta ,\) \(y=r\ \text{sin}\ \theta ,\) and \(z=z.\) The Jacobian for the transformation is
       \[\begin{array}{llllllllllllll}J(r,\theta ,z) & =\frac{∂(x,y,z)}{∂(r,\theta ,z)}=|\begin{array}{lllll}\frac{∂x}{∂r} & & \frac{∂x}{∂\theta } & & \frac{∂x}{∂z} \\ \frac{∂y}{∂r} & & \frac{∂y}{∂\theta } & & \frac{∂y}{∂z} \\ \frac{∂z}{∂r} & & \frac{∂z}{∂\theta } & & \frac{∂z}{∂z}\end{array}| \\ & =|\begin{array}{lllll}\text{cos}\ \theta & & -r\ \text{sin}\ \theta & & 0 \\ \text{sin}\ \theta & & r\ \text{cos}\ \theta & & 0 \\ 0 & & 0 & & 1\end{array}|=r\ {\text{cos}}^{2}\theta +r\ {\text{sin}}^{2}\theta =r({\text{cos}}^{2}\theta +{\text{sin}}^{2}\theta )=r.\end{array}\]
       We know that \(r\ge 0,\) so \(|J(r,\theta ,z)|=r.\) Then the triple integral is
       \[\underset{D}{∭}f(x,y,z)dV=\underset{G}{∭}f(r\ \text{cos}\ \theta ,r\ \text{sin}\ \theta ,z)r\ dr\ d\theta \ dz.\]
       
    2. For spherical coordinates, the transformation is \(T(\rho ,\theta ,\phi )=(x,y,z)\) from the Cartesian \(p\theta \phi \text{-plane}\) to the Cartesian \(xyz\text{-plane}\) (). Here \(x=\rho \ \text{sin}\ \phi \ \text{cos}\ \theta ,\) \(y=\rho \ \text{sin}\ \phi \ \text{sin}\ \theta ,\) and \(z=\rho \ \text{cos}\ \phi .\) The Jacobian for the transformation is
       \[J(\rho ,\theta ,\phi )=\frac{∂(x,y,z)}{∂(\rho ,\theta ,\phi )}=|\begin{array}{lllll}\frac{∂x}{∂\rho } & & \frac{∂x}{∂\theta } & & \frac{∂x}{∂\phi } \\ \frac{∂y}{∂\rho } & & \frac{∂y}{∂\theta } & & \frac{∂y}{∂\phi } \\ \frac{∂z}{∂\rho } & & \frac{∂z}{∂\theta } & & \frac{∂z}{∂\phi }\end{array}|=|\begin{array}{lllll}\text{sin}\ \phi \ \text{cos}\ \theta & & \rho \ \text{sin}\ \phi \ \text{sin}\ \theta & & \rho \ \text{cos}\ \phi \ \text{cos}\ \theta \\ \text{sin}\ \phi \ \text{sin}\ \theta & & \rho \ \text{sin}\ \phi \ \text{cos}\ \theta & & \rho \ \text{cos}\ \phi \ \text{sin}\ \theta \\ \text{cos}\ ϕ & & 0 & & -\rho \ \text{sin}\ \phi \end{array}|.\]
       Expanding the determinant with respect to the third row:
       \[\begin{array}{lllllllll} \\ \\ =\text{cos}\ \phi |\begin{array}{lll}-\rho \ \text{sin}\ \phi \ \text{sin}\ \theta & & \rho \ \text{cos}\ \phi \ \text{cos}\ \theta \\ \rho \ \text{sin}\ \phi \ \text{sin}\ \theta & & \rho \ \text{cos}\ \phi \ \text{sin}\ \theta \end{array}|-\rho \ \text{sin}\ \phi |\begin{array}{lll}\text{sin}\ \phi \ \text{cos}\ \theta & & -\rho \ \text{sin}\ \phi \ \text{sin}\ \theta \\ \text{sin}\ \phi \ \text{sin}\ \theta & & \rho \ \text{sin}\ \phi \ \text{cos}\ \theta \end{array}| \\ =\text{cos}\ \phi (\text{-}{\rho }^{2}\text{sin}\ \phi \ \text{cos}\ \phi \ {\text{sin}}^{2}\theta -{\rho }^{2}\text{sin}\ \phi \ \text{cos}\ \phi \ {\text{cos}}^{2}\theta ) \\ -\rho \ \text{sin}\ \phi (\rho \ {\text{sin}}^{2}\phi \ {\text{cos}}^{2}\theta +\rho \ {\text{sin}}^{2}\phi \ {\text{sin}}^{2}\theta ) \\ =\text{-}{\rho }^{2}\text{sin}\ \phi \ {\text{cos}}^{2}\phi ({\text{sin}}^{2}\theta +{\text{cos}}^{2}\theta )-{\rho }^{2}\text{sin}\ \phi \ {\text{sin}}^{2}\phi ({\text{sin}}^{2}\theta +{\text{cos}}^{2}\theta ) \\ =\text{-}{\rho }^{2}\text{sin}\ \phi \ {\text{cos}}^{2}\phi -{\rho }^{2}\text{sin}\ \phi \ {\text{sin}}^{2}\phi \\ =\text{-}{\rho }^{2}\text{sin}\ \phi ({\text{cos}}^{2}\phi +{\text{sin}}^{2}\phi )=\text{-}{\rho }^{2}\text{sin}\ \phi .\end{array}\]
       Since \(0\le \phi \le \pi ,\) we must have \(\text{sin}\ \phi \ge 0.\) Thus \(|J(\rho ,\theta ,\phi )|=|\text{-}{\rho }^{2}\text{sin}\ \phi |={\rho }^{2}\text{sin}\ \phi .\)
       Then the triple integral becomes
       \[\underset{D}{∭}f(x,y,z)dV=\underset{G}{∭}f(\rho \ \text{sin}\ \phi \ \text{cos}\ \theta ,\rho \ \text{sin}\ \phi \ \text{sin}\ \theta ,\rho \ \text{cos}\ \phi ){\rho }^{2}\text{sin}\ \phi \ d\rho \ d\phi \ d\theta .\]

14. Evaluate the triple integral

    \[\int _{0}^{3}\ \int _{0}^{4}\ \int _{y\text{/}2}^{(y\text{/}2)+1}(x+\frac{z}{3})dx\ dy\ dz\]

    in \(xyz\text{-space}\) by using the transformation

    \[u=(2x-y)\text{/}2,v=y\text{/}2,\text{and}\ w=z\text{/}3.\]

    Then integrate over an appropriate region in \(uvw\text{-space}\text{.}\)

    Jiżvelaw it-tweġiba

    As before, some kind of sketch of the region \(G\) in \(xyz\text{-space}\) over which we have to perform the integration can help identify the region \(D\) in \(uvw\text{-space}\) (). Clearly \(G\) in \(xyz\text{-space}\) is bounded by the planes \(x=y\text{/}2,x=(y\text{/}2)+1,y=0,\) \(y=4,\) \(z=0,\text{and}\ z=4.\) We also know that we have to use \(u=(2x-y)\text{/}2,v=y\text{/}2,\text{and}\ w=z\text{/}3\) for the transformations. We need to solve for \(x,y,\text{and}\ z.\) Here we find that \(x=u+v,\) \(y=2v,\) and \(z=3w.\)

    Using elementary algebra, we can find the corresponding surfaces for the region \(G\) and the limits of integration in \(uvw\text{-space}\text{.}\) It is convenient to list these equations in a table.

    Equations in \(xyz\) for the region \(D\)	Corresponding equations in \(uvw\) for the region \(G\)	Limits for the integration in \(uvw\)
    \(x=y\text{/}2\)	\(u+v=2v\text{/}2=v\)	\(u=0\)
    \(x=y\text{/}2\)	\(u+v=(2v\text{/}2)+1=v+1\)	\(u=1\)
    \(y=0\)	\(2v=0\)	\(v=0\)
    \(y=4\)	\(2v=4\)	\(v=2\)
    \(z=0\)	\(3w=0\)	\(w=0\)
    \(z=3\)	\(3w=3\)	\(w=1\)

    Now we can calculate the Jacobian for the transformation:

    \[J(u,v,w)=|\begin{array}{lllll}\frac{∂x}{∂u} & & \frac{∂x}{∂v} & & \frac{∂x}{∂w} \\ \frac{∂y}{∂u} & & \frac{∂y}{∂v} & & \frac{∂y}{∂w} \\ \frac{∂z}{∂u} & & \frac{∂z}{∂v} & & \frac{∂z}{∂w}\end{array}|=|\begin{array}{lllll}1 & & 1 & & 0 \\ 0 & & 2 & & 0 \\ 0 & & 0 & & 3\end{array}|=6.\]

    The function to be integrated becomes

    \[f(x,y,z)=x+\frac{z}{3}=u+v+\frac{3w}{3}=u+v+w.\]

    We are now ready to put everything together and complete the problem.

    \[\begin{array}{l} \\ \\ \\ \\ \int _{0}^{3}\ \int _{0}^{4}\ \int _{y\text{/}2}^{(y\text{/}2)+1}(x+\frac{z}{3})dx\ dy\ dz \\ \\ =\int _{0}^{1}\ \int _{0}^{2}\ \int _{0}^{1}(u+v+w)|J(u,v,w)|du\ dv\ dw=\int _{0}^{1}\ \int _{0}^{2}\ \int _{0}^{1}(u+v+w)|6|du\ dv\ dw \\ =6\int _{0}^{1}\ \int _{0}^{2}\ \int _{0}^{1}(u+v+w)du\ dv\ dw=6{\int _{0}^{1}\ \int _{0}^{2}[\frac{{u}^{2}}{2}+vu+wu]}_{0}^{1}dv\ dw \\ =6\int _{0}^{1}\ \int _{0}^{2}(\frac{1}{2}+v+w)dv\ dw=6{\int _{0}^{1}[\frac{1}{2}v+\frac{{v}^{2}}{2}+wv]}_{0}^{2}dw \\ =6\int _{0}^{1}(3+2w)dw=6{[3w+{w}^{2}]}_{0}^{1}=24.\end{array}\]

15. Let \(D\) be the region in \(xyz\text{-space}\) defined by \(1\le x\le 2,0\le xy\le 2,\text{and}\ 0\le z\le 1.\)

    Evaluate \(\underset{D}{∭}({x}^{2}y+3xyz)dx\ dy\ dz\) by using the transformation \(u=x,v=xy,\) and \(w=3z.\)

    Jiżvelaw it-tweġiba

    \(\int _{0}^{3}\ \int _{0}^{2}\ \int _{1}^{2}(\frac{v}{3}+\frac{vw}{3u})du\ dv\ dw=2+\text{ln}\ 8\)

16. \(x=2u,y=3v\)

17. \(x=\frac{u}{2},y=\frac{v}{3}\)

    Jiżvelaw it-tweġiba

    a. \(T(u,v)=(g(u,v),h(u,v)),x=g(u,v)=\frac{u}{2}\) and \(y=h(u,v)=\frac{v}{3}.\) The functions \(g\) and \(h\) are continuous and differentiable, and the partial derivatives \({g}_{u}(u,v)=\frac{1}{2},\) \({g}_{v}(u,v)=0,{h}_{u}(u,v)=0\ \text{and}\ {h}_{v}(u,v)=\frac{1}{3}\) are continuous on \(S;\) b. \(T(0,0)=(0,0),\) \(T(1,0)=(\frac{1}{2},0),T(0,1)=(0,\frac{1}{3}),\) and \(T(1,1)=(\frac{1}{2},\frac{1}{3});\) c. \(R\) is the rectangle of vertices \((0,0),(\frac{1}{2},0),(\frac{1}{2},\frac{1}{3}),\text{and}\ (0,\frac{1}{3})\) in the \(xy\text{-plane;}\) the following figure.
    

18. \(x=u-v,y=u+v\)

19. \(x=2u-v,y=u+2v\)

    Jiżvelaw it-tweġiba

    a. \(T(u,v)=(g(u,v),h(u,v)),x=g(u,v)=2u-v,\) and \(y=h(u,v)=u+2v.\) The functions \(g\) and \(h\) are continuous and differentiable, and the partial derivatives \({g}_{u}(u,v)=2,\) \({g}_{v}(u,v)=-1,\) \({h}_{u}(u,v)=1,\) and \({h}_{v}(u,v)=2\) are continuous on \(S;\) b. \(T(0,0)=(0,0),\) \(T(1,0)=(2,1),\) \(T(0,1)=(-1,2),\) and \(T(1,1)=(1,3);\) c. \(R\) is the sqaure of vertices \((0,0),(2,1),(1,3),\text{and}\ (-1,2)\) in the \(xy\text{-plane;}\) see the following figure.
    

20. \(x={u}^{2},y={v}^{2}\)

21. \(x={u}^{3},y={v}^{3}\)

    Jiżvelaw it-tweġiba

    a. \(T(u,v)=(g(u,v),h(u,v)),x=g(u,v)={u}^{3},\) and \(y=h(u,v)={v}^{3}.\) The functions \(g\) and \(h\) are continuous and differentiable, and the partial derivatives \({g}_{u}(u,v)=3{u}^{2},\) \({g}_{v}(u,v)=0,\) \({h}_{u}(u,v)=0,\) and \({h}_{v}(u,v)=3{v}^{2}\) are continuous on \(S;\) b. \(T(0,0)=(0,0),\) \(T(1,0)=(1,0),\) \(T(0,1)=(0,1),\) and \(T(1,1)=(1,1);\) c. \(R\) is the unit square in the \(xy\text{-plane;}\) see the following figure.

22. \(x={u}^{2},y={v}^{2},\text{where}\ S\) is the rectangle of vertices \((-1,0),(1,0),(1,1),\text{and}\ (-1,1).\)

23. \(x={u}^{4},y={u}^{2}+v,\text{where}\ S\) is the triangle of vertices \((-2,0),(2,0),\text{and}\ (0,2).\)

    Jiżvelaw it-tweġiba

    \(T\) is not one-to-one: two points of \(S\) have the same image. Indeed, \(T(-2,0)=T(2,0)=(16,4).\)

24. \(x=2u,y=3v,\text{where}\ S\) is the square of vertices \((-1,1),(-1,-1),(1,-1),\text{and}\ (1,1).\)

25. \(T(u,v)=(2u-v,u),\) where \(S\) is the triangle of vertices \((-1,1),(-1,-1),\text{and}\ (1,-1).\)

    Jiżvelaw it-tweġiba

    \(T\) is one-to-one: We argue by contradiction. \(T({u}_{1},{v}_{1})=T({u}_{2},{v}_{2})\) implies \(2{u}_{1}-{v}_{1}=2{u}_{2}-{v}_{2}\) and \({u}_{1}={u}_{2}.\) Thus, \({u}_{1}={u}_{2}\) and \({v}_{1}={v}_{2}.\)

26. \(x=u+v+w,y=u+v,z=w,\) where \(S=R={ℝ}^{3}.\)

27. \(x={u}^{2}+v+w,y={u}^{2}+v,z=w,\) where \(S=R={ℝ}^{3}.\)

    Jiżvelaw it-tweġiba

    \(T\) is not one-to-one: \(T(1,v,w)=(-1,v,w)\)

28. \(x=4u,y=5v,\) where \(S=R={ℝ}^{2}.\)

29. \(x=u+2v,y=\text{-}u+v,\) where \(S=R={ℝ}^{2}.\)

    Jiżvelaw it-tweġiba

    \(u=\frac{x-2y}{3},v=\frac{x+y}{3}\)

30. \(x={e}^{2u+v},y={e}^{u-v},\) where \(S={ℝ}^{2}\) and \(R=\{(x,y)|x>0,y>0\}\)

31. \(x=\text{ln}\ u,y=\text{ln}(uv),\) where \(S=\{(u,v)|u>0,v>0\}\) and \(R={ℝ}^{2}.\)

    Jiżvelaw it-tweġiba

    \(u={e}^{x},v={e}^{\text{-}x+y}\)

32. \(x=u+v+w,y=3v,z=2w,\) where \(S=R={ℝ}^{3}.\)

33. \(x=u+v,y=v+w,z=u+w,\) where \(S=R={ℝ}^{3}.\)

    Jiżvelaw it-tweġiba

    \(u=\frac{x-y+z}{2},v=\frac{x+y-z}{2},w=\frac{\text{-}x+y+z}{2}\)

34. \(x=au,y=bv,R=\{(x,y)|{x}^{2}+{y}^{2}\le {a}^{2}{b}^{2}\},\) where \(a,b>0\)

35. \(x=au,y=bv,R=\{(x,y)|\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}\le 1\},\) where \(a,b>0\)

    Jiżvelaw it-tweġiba

    \(S=\{(u,v)|{u}^{2}+{v}^{2}\le 1\}\)

36. \(x=\frac{u}{a},y=\frac{v}{b},z=\frac{w}{c},\) \(R=\{(x,y)|{x}^{2}+{y}^{2}+{z}^{2}\le 1\},\) where \(a,b,c>0\)

37. \(x=au,y=bv,z=cw,R=\{(x,y)|\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}-\frac{{z}^{2}}{{c}^{2}}\le 1,z>0\},\) where \(a,b,c>0\)

    Jiżvelaw it-tweġiba

    \(R=\{(u,v,w)|{u}^{2}-{v}^{2}-{w}^{2}\le 1,w>0\}\)

38. \(x=u+2v,y=\text{-}u+v\)

39. \(x=\frac{{u}^{3}}{2},y=\frac{v}{{u}^{2}}\)

    Jiżvelaw it-tweġiba

    \(\frac{3}{2}\)

40. \(x={e}^{2u-v},y={e}^{u+v}\)

Ipprova tiegħek stess

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.

Aktar fil Multivariable Calculus

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