Section 5.7Change of Variables in Multiple Integrals

Since \(r\) varies from 0 to 1 in the \(rθ\text{-plane},\) we have a circular disc of radius 0 to 1 in the \(xy\text{-plane}\text{.}\) Because \(θ\) varies from 0 to \(\pi \text{/2}\) in the \(rθ\text{-plane},\) we end up getting a quarter circle of radius \(1\) in the first quadrant of the \(xy\text{-plane}\) (Figure 2). 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 \(T({r}_{1},{θ}_{1})=T({r}_{2},{θ}_{2})\) and show as a consequence that \(({r}_{1},{θ}_{1})=({r}_{2},{θ}_{2}).\) In this case, we have

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

Dividing, we obtain

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

since the cotangent function is one-one function in the interval \(0\le θ\le \pi \text{/}2.\) Also, since \(0<r\le 1,\) we have \({r}_{1}={r}_{2},{θ}_{1}={θ}_{2}.\) Therefore, \(({r}_{1},{θ}_{1})=({r}_{2},{θ}_{2})\) and \(T\) is a one-to-one transformation from \(G\) into \(R.\)

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

The triangle and its image are shown in Figure 3. 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 \(A\prime\) 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 \({B}^{\prime }\) 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 \({C}^{\prime }\) 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 \(xy\text{-plane}\text{.}\)

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

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

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

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

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 \(r=2\,\text{cos}\,θ\) so the region of integration in polar coordinates is bounded by \(0\le r\le 2 \text{cos}\,θ\) and \(0\le θ\le \frac{\pi }{2}.\)

The Jacobian is \(J(r,θ)=r,\) as shown in Example 3. Since \(r\ge 0,\) we have \(|J(r,θ)|=r.\)

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

\[∫02\,∫0\sqrt{2x-{x}^{2}}\sqrt{{x}^{2}+{y}^{2}}dy\,dx=∫0\pi \text{/}2\,∫02\,\text{cos}\,θr|J(r,θ)|dr\,dθ=∫0\pi \text{/}2\,∫02\,\text{cos}\,θ{r}^{2}dr\,dθ.\]

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are \(x-y+1=0,x-y-1=0,\) \(x-3y+5=0,\text{and}\,x-3y+9=0\) (Figure 8). 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

\[\underset{R}{∬}(x-y)dy\,dx=∫-9-5\,∫-11J(u,v)u\,du\,dv=∫-9-5\,∫-11(\frac{1}{2})u\,du\,dv,\]

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.

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 (Figure 9).

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 Figure 9) 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

\[\underset{R}{∬}(x-y){e}^{{x}^{2}-{y}^{2}}dA=\frac{1}{2}∫13\,∫-1\text{/}v1\text{/}vu{e}^{uv}du\,dv.\]

Doing the evaluation, we have

\[\frac{1}{2}∫13\,∫-1\text{/}v1\text{/}vu{e}^{uv}du\,dv=\frac{2}{3e}\approx 0.245.\]

• For cylindrical coordinates, the transformation is \(T(r,θ,z)=(x,y,z)\) from the Cartesian \(rθz\text{-plane}\) to the Cartesian \(xyz\text{-plane}\) (Figure 11). Here \(x=r\,\text{cos}\,θ,\) \(y=r\,\text{sin}\,θ,\) and \(z=z.\) The Jacobian for the transformation is
  
  We know that \(r\ge 0,\) so \(|J(r,θ,z)|=r.\) Then the triple integral is
  
  

  \[\begin{array}{ll}J(r,θ,z) & =\frac{∂(x,y,z)}{∂(r,θ,z)}=|\begin{array}{lllll}\frac{∂x}{∂r} & & \frac{∂x}{∂θ} & & \frac{∂x}{∂z} \\ \frac{∂y}{∂r} & & \frac{∂y}{∂θ} & & \frac{∂y}{∂z} \\ \frac{∂z}{∂r} & & \frac{∂z}{∂θ} & & \frac{∂z}{∂z}\end{array}| \\ & =|\begin{array}{lllll}\text{cos}\,θ & & -r\,\text{sin}\,θ & & 0 \\ \text{sin}\,θ & & r\,\text{cos}\,θ & & 0 \\ 0 & & 0 & & 1\end{array}|=r\,{\text{cos}}^{2}θ+r\,{\text{sin}}^{2}θ=r({\text{cos}}^{2}θ+{\text{sin}}^{2}θ)=r.\end{array}\]

  \[\underset{D}{∭}f(x,y,z)dV=\underset{G}{∭}f(r\,\text{cos}\,θ,r\,\text{sin}\,θ,z)r\,dr\,dθ\,dz.\]

  

• For spherical coordinates, the transformation is \(T(ρ,θ,φ)=(x,y,z)\) from the Cartesian \(pθφ\text{-plane}\) to the Cartesian \(xyz\text{-plane}\) (Figure 12). Here \(x=ρ\,\text{sin}\,φ\,\text{cos}\,θ,\) \(y=ρ\,\text{sin}\,φ\,\text{sin}\,θ,\) and \(z=ρ\,\text{cos}\,φ.\) The Jacobian for the transformation is
  
  Expanding the determinant with respect to the third row:
  
  Since \(0\le φ\le \pi ,\) we must have \(\text{sin}\,φ\ge 0.\) Thus \(|J(ρ,θ,φ)|=|\text{-}{ρ}^{2}\text{sin}\,φ|={ρ}^{2}\text{sin}\,φ.\)
  Then the triple integral becomes
  

  \[J(ρ,θ,φ)=\frac{∂(x,y,z)}{∂(ρ,θ,φ)}=|\begin{array}{lllll}\frac{∂x}{∂ρ} & & \frac{∂x}{∂θ} & & \frac{∂x}{∂φ} \\ \frac{∂y}{∂ρ} & & \frac{∂y}{∂θ} & & \frac{∂y}{∂φ} \\ \frac{∂z}{∂ρ} & & \frac{∂z}{∂θ} & & \frac{∂z}{∂φ}\end{array}|=|\begin{array}{lllll}\text{sin}\,φ\,\text{cos}\,θ & & ρ\,\text{sin}\,φ\,\text{sin}\,θ & & ρ\,\text{cos}\,φ\,\text{cos}\,θ \\ \text{sin}\,φ\,\text{sin}\,θ & & ρ\,\text{sin}\,φ\,\text{cos}\,θ & & ρ\,\text{cos}\,φ\,\text{sin}\,θ \\ \text{cos}\,ϕ & & 0 & & -ρ\,\text{sin}\,φ\end{array}|.\]

  \[\begin{array}{l} \\ \\ =\text{cos}\,φ|\begin{array}{lll}-ρ\,\text{sin}\,φ\,\text{sin}\,θ & & ρ\,\text{cos}\,φ\,\text{cos}\,θ \\ ρ\,\text{sin}\,φ\,\text{sin}\,θ & & ρ\,\text{cos}\,φ\,\text{sin}\,θ\end{array}|-ρ\,\text{sin}\,φ|\begin{array}{lll}\text{sin}\,φ\,\text{cos}\,θ & & -ρ\,\text{sin}\,φ\,\text{sin}\,θ \\ \text{sin}\,φ\,\text{sin}\,θ & & ρ\,\text{sin}\,φ\,\text{cos}\,θ\end{array}| \\ =\text{cos}\,φ(\text{-}{ρ}^{2}\text{sin}\,φ\,\text{cos}\,φ\,{\text{sin}}^{2}θ-{ρ}^{2}\text{sin}\,φ\,\text{cos}\,φ\,{\text{cos}}^{2}θ) \\ \,-ρ\,\text{sin}\,φ(ρ\,{\text{sin}}^{2}φ\,{\text{cos}}^{2}θ+ρ\,{\text{sin}}^{2}φ\,{\text{sin}}^{2}θ) \\ =\text{-}{ρ}^{2}\text{sin}\,φ\,{\text{cos}}^{2}φ({\text{sin}}^{2}θ+{\text{cos}}^{2}θ)-{ρ}^{2}\text{sin}\,φ\,{\text{sin}}^{2}φ({\text{sin}}^{2}θ+{\text{cos}}^{2}θ) \\ =\text{-}{ρ}^{2}\text{sin}\,φ\,{\text{cos}}^{2}φ-{ρ}^{2}\text{sin}\,φ\,{\text{sin}}^{2}φ \\ =\text{-}{ρ}^{2}\text{sin}\,φ({\text{cos}}^{2}φ+{\text{sin}}^{2}φ)=\text{-}{ρ}^{2}\text{sin}\,φ.\end{array}\]

  

  \[\underset{D}{∭}f(x,y,z)dV=\underset{G}{∭}f(ρ\,\text{sin}\,φ\,\text{cos}\,θ,ρ\,\text{sin}\,φ\,\text{sin}\,θ,ρ\,\text{cos}\,φ){ρ}^{2}\text{sin}\,φ\,dρ\,dφ\,dθ.\]

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 \(uvw\text{-space}\) (Figure 13). 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.

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.

\[\begin{array}{l} \\ \\ \\ \\ \,∫03\,∫04\,∫y\text{/}2(y\text{/}2)+1(x+\frac{z}{3})dx\,dy\,dz \\ \\ =∫01\,∫02\,∫01(u+v+w)|J(u,v,w)|du\,dv\,dw=∫01\,∫02\,∫01(u+v+w)|6|du\,dv\,dw \\ =6∫01\,∫02\,∫01(u+v+w)du\,dv\,dw=6{∫01\,∫02[\frac{{u}^{2}}{2}+vu+wu]}_{0}^{1}dv\,dw \\ =6∫01\,∫02(\frac{1}{2}+v+w)dv\,dw=6{∫01[\frac{1}{2}v+\frac{{v}^{2}}{2}+wv]}_{0}^{2}dw \\ =6∫01(3+2w)dw=6{[3w+{w}^{2}]}_{0}^{1}=24.\end{array}\]