Change of Variables in Multiple Integrals

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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

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   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 \((1,1).\)

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   The triangle and its image are 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 \({T}^{-1}(x,y).\)

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

4. Find the Jacobian of the transformation

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   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

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   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: \(T(u,v)=(u+v,2v).\)

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

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.

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   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 \(J(r,\theta )=r,\) 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.

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

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.

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   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.

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    \(x=\frac{1}{2}(v+u)\) 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 ).

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    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 \(y=\sqrt{x},x=2,\text{and}\ y=0.\)

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

13. Derive the formula in triple integrals for

    1. cylindrical and
    2. spherical coordinates.
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    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 \(uvw\text{-space}\text{.}\)

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    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 \(w=3z.\)

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

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

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

    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 \((-1,0),(1,0),(1,1),\text{and}\ (-1,1).\)

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

    \(T\) 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 \((-1,1),(-1,-1),(1,-1),\text{and}\ (1,1).\)

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

    \(T\) 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 \(S=R={ℝ}^{3}.\)

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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 \(a,b,c>0\)

    Kerjakan langkah demi langkah Petunjuk

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 \(a,b,c>0\)

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

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

    Kerjakan langkah demi langkah Petunjuk

    Tampilkan jawaban

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

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

    Kerjakan langkah demi langkah Petunjuk

Bagian dari halaman ini diadaptasi dari OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Diringkas dan dijelaskan ulang di sini; kesalahan adalah tanggung jawab kami. Karena sumbernya dilisensikan untuk penggunaan non-komersial, pelajaran ini gratis, menjual apa-apa, dan dibagi di bawah lisensi yang sama.

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