Change of Variables in Multiple Integrals

We omit the proof - you’ll do quite fine in calculus without knowing this proof.

The transformation \((x,y)\mapsto (2x+3y,x-3y)\) is linear, and the image of the original parallelogram must be parallelogram as well. The vertex \((0,0)\) gets mapped to the vertex \((0,0)\) of the new parallelogram. Similarly, \(\left(1,\frac13\right)\) gets mapped to \((3,0)\), \(\left(\frac43,\frac19\right)\mapsto (3,1)\), and \(\left(\frac13,-\frac29\right)\mapsto (0,1)\).

In order to find the Jacobian we need to express \(x\) and \(y\) in terms of \(u\) and \(v\). We do this by solving the system \(u=x+3y\), \(v=y\), and we get:

\begin{eqnarray*} x&=&\frac{u+v}3\newline y&=&\frac{u-2v}9.\end{eqnarray*} The Jacobian can be found as: \[\frac{\partial(x,y)}{\partial{u,v}}=\det\left|\begin{array}{cc}\frac13&\frac13\newline \frac19&-\frac29\end{array}\right|=-\frac19.\]

Denote by \(E\) the parallelogram with vertices \((0,0)\), \((3,0)\), \((3,1)\), and \((0,1)\). We have \begin{eqnarray*}\iint_D e^{2x+3y}\cdot \cos(x-3y)\,dxdy&=&\iint_E e^u\cdot \cos v \cdot\left|-\frac19\right|\,dudv=\frac19\iint_Ee^u\cos v\,dudv \newline &=&\frac19\int_0^1\int_0^3e^u\cos v\,dudv=\frac19 \int_0^1\left.\left(\cos v\cdot e^u\right)\right|_{u=0}^{u=3}\,dv\newline &=&\frac19\int_0^1 \left(e^3-1\right)\cdot \cos v\,dv=\frac{e^3-1}9\cdot \left.\sin v\right|_{v=0}^{v=1}\newline &=&\frac{\left(e^3-1\right)\cdot \sin 1}{9}.\end{eqnarray*}

Omitted. See any real analysis textbook.

Let us use the following substitution: \begin{eqnarray*} x&=&r\cos\theta\newline y&=&r\sin\theta\newline 0\leq &r&\leq 3\newline 0\leq &\theta &< 2\pi. \end{eqnarray*} The transformation \((r,\theta)\mapsto (r\cos\theta,r\sin\theta)\) is a bijection between the rectangle \([0,3]\times[0,2\pi]\) in the \((r,\theta)\)-plane and the disc of radius \(3\).

Since \(x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\), the integral becomes: \begin{eqnarray*} \iint_D \cos\left(x^2+y^2\right)\,dxdy&=&\int_0^{2\pi}\int_0^3 \cos\left(r^2\right) \cdot r\,drd\theta. \end{eqnarray*} For the integral \(\int_0^3\cos\left(r^2\right)r\,dr\) we use the substitution \(r^2=u\). Then we have \(r=\sqrt u\) and \(dr=\frac1{2\sqrt u}\,du\). The bounds of integration become \(0\leq u\leq 9\), and the integral is \[\int_0^3\cos(r^2)r\,dr=\int_0^9\cos u\cdot \sqrt u\cdot \frac1{2\sqrt u}\,du=\frac12\int_0^9\cos u\,du=\frac{\sin 9}2.\] Therefore, \[ \iint_D \cos\left(x^2+y^2\right)\,dxdy= \int_0^{2\pi}\frac{\sin9}2\,d\theta= \sin 9\cdot \pi.\]

Denote by \(E(a,b)\) the region enclosed by the ellipse. The area of the ellipse is given by the equation \(A=\iint_{E(a,b)}1\,dxdy\). We make a substitution \(x=ar\cos \theta\), \(y=br\sin \theta\), \(0\leq r\leq 1\), \(0\leq \theta < 2\pi\). The Jacobian is \[J=\left| \mbox{det }\left[\begin{array}{cc}a\cos\theta&-ar\sin\theta\newline b\sin\theta&br\cos\theta \end{array} \right]\right|=abr.\] Then \(A=\int_0^{2\pi}\int_0^1abr\,drd\theta=ab\pi\).

Let us use the following substitution: \begin{eqnarray*} x&=&r\cos\theta\newline y&=&r\sin\theta\newline z&=&z\newline 0\leq &r&\leq 2\newline 0\leq &\theta &< \frac{\pi}4\newline 0\leq&z&\leq 4-r^2. \end{eqnarray*} The transformation \((r,\theta,z)\mapsto (r\cos\theta,r\sin\theta,z)\) is a bijection between the solid the solid \(B\) defined as: \[B=\left\{(r,\theta,z):0\leq r\leq 2, 0\leq \theta\leq \frac{\pi}4, 0\leq z \leq 4-r^2\right\}\] in the \((r,\theta,z)\)-space and the solid \(D\) from the formulation of the problem.

Since \(x^2+y^2=r^2\cos^2\theta+r^2\sin^2\theta=r^2\), the integral becomes: \begin{eqnarray*} \iiint_D e^{x^2+y^2}\,dxdydz&=&\int_0^{2}\int_0^{\frac{\pi}4}\int_0^{4-r^2} e^{r^2} \cdot r\,dzd\theta dr\newline &=&\int_0^{2}\int_0^{\frac{\pi}4} e^{r^2} \cdot r(4-r^2)\,d\theta dr\newline &=&\frac{\pi}4\cdot\int_0^2 r(4-r^2)e^{r^2}\,dr. \end{eqnarray*} In the last integral we use the substitution \(r^2=u\). Then we have \(r=\sqrt u\) and \(dr=\frac1{2\sqrt u}\,du\). The bounds of integration become \(0\leq u\leq 4\), and the integral is \[\int_0^2 r(4-r^2)e^{r^2}\,dr=\int_0^4\sqrt u\cdot(4-u)\cdot e^u\cdot\frac1{2\sqrt u}\,du=\frac12\int_0^44e^u\,du-\frac12\int_0^4ue^u\,du.\] The first term on the right-hand side is equal to \(2\left(e^4-1\right)\), and for the second we use the integration by parts. We take \(f=u\), \(dg=e^udu\), which gives us \(g=e^u\) and the integral becomes: \[\int_0^4ue^u\,du=\left.ue^u\right|_0^4-\int_0^4e^u\,du=4e^4-e^4+1.\] Therefore \[\int_0^2r(4-r^2)e^{r^2}\,dr=2e^4-2-2e^4+\frac{e^4}2-\frac12=\frac{e^4-5}2,\] thus \[ \iiint_D e^{x^2+y^2}\,dxdydz= \frac{\left(e^4-5\right)\pi}8. \]

Let us use the following substitution: \begin{eqnarray*} x&=&\rho\cos\theta\sin\phi\newline y&=&\rho\sin\theta\sin\phi\newline z&=&\rho\cos\phi\newline 0\leq &\rho&\leq 3\newline 0\leq &\theta &< \frac{\pi}4\newline 0\leq&\phi&\leq \frac{\pi}2. \end{eqnarray*} The evaluation of the integral is now easy as it becomes an iterated integral in variables \(\rho\), \(\theta\), and \(\phi\). \begin{eqnarray*} \iiint_D e^{\sqrt{x^2+y^2+z^2}}\,dV&=& \int_0^{3}\int_0^{\frac{\pi}4}\int_0^{\frac{\pi}2} e^{\rho}\cdot \rho^2\cdot\sin\phi\,d\phi d\theta d\rho =\int_0^3\int_0^{\frac{\pi}4}e^{\rho}\cdot \rho^2\cdot \left.\left(-\cos\phi\right)\right|_{\phi=0}^{\phi=\frac{\pi}2}\,d\theta d\rho\newline &=&\int_0^3\int_0^{\frac{\pi}4}e^{\rho}\cdot \rho^2\cdot 1\,d\theta d\rho=\frac{\pi}4\cdot\int_0^3 \rho^2e^{\rho}\,d\rho.\end{eqnarray*} In the last integral we can use integration by parts with functions \(u=\rho^2\), \(d v=e^{\rho}\,d\rho\). Then we have \(du=2\rho\,d\rho\) and we can take \(v=e^{\rho}\). The integral becomes: \begin{eqnarray*} \int_0^3e^{\rho}\rho^2\,d\rho=\left.\rho^2e^{\rho}\right|_0^3-2\int_0^3\rho e^{\rho}\,d\rho &=& 9e^3-\left.2\rho e^{\rho}\right|0^{3}+2\int_0^3e^{\rho}\,d\rho=9e^3-6e^3+2e^3-2=5e^3-2. \end{eqnarray*} Thus the final result is: \begin{eqnarray*} \iiint_D e^{\sqrt{x^2+y^2+z^2}}\,dV&=&\frac{\pi\cdot\left(5e^3-2\right)}4. \end{eqnarray*}

We will use the substitution \(x=u^2\), \(y=v^2\), \(z=w^2\). Then the variables \(u\), \(v\), and \(w\) satisfy \(u,v,w\geq 0\) and \(u+v+w\leq 1\). The Jacobian can be easily calculated: \[J=\left| \mbox{det }\left[ \begin{array}{ccc}2u&0&0\newline 0&2v&0\newline 0&0&2w\end{array} \right] \right|=|8uvw|=8uvw.\] The integral now becomes \begin{eqnarray*}\iiint_S\frac1{\sqrt{xyz}}\,dxdydz&=&\int_0^1\int_0^{1-w}\int_0^{1-v-w} \frac{8uvw}{uvw}\,dudvdw\\&=& 8\int_0^1\int_0^{1-w}(1-v-w)\,dvdw\\&=& 8\int_0^1\frac12(1-w)^2\,dw=\frac43. \end{eqnarray*}

Second solution. We can evaluate this integral without using a substitution. \begin{eqnarray*} \iiint_S\frac1{\sqrt{xyz}}\,dxdydz&=&\int_0^1\int_0^{(1-\sqrt z)^2}\int_0^{(1-\sqrt z-\sqrt y)^2}\frac1{\sqrt{yz}}\cdot \frac1{\sqrt x}\,dxdydz\\&=& \int_0^1\int_0^{(1-\sqrt z)^2}\frac2{\sqrt{yz}}\cdot (1-\sqrt y-\sqrt z)\,dydz\\ &=&2\int_0^1\frac1{\sqrt z}\int_0^{(1-\sqrt z)^2}\left(\frac{1-\sqrt z}{\sqrt y}-1\right)\,dydz\\ &=&2\int_0^1\frac1{\sqrt z}\frac{(1-\sqrt z)^2}{\sqrt z}\,dz\\&=& 2\int_0^1\left(\frac1{\sqrt z}-2+\sqrt z\right)\,dz\\&=&2\cdot\left(2-2+\frac23\right)=\frac43. \end{eqnarray*}

Let \(A\) be the triangle with vertices \((0,0)\), \((1,1)\), and \((0,1)\). Let \(B=\{(x,y):x^2+y^2\leq 2, 0\leq x, y\leq x\}\). Since the integrand is symmetric with respect to \(x\) and \(y\) axis, we have: \[\iint_D (x^2+y^2)\,dA=2\left(\iint_A (x^2+y^2)\, dA+\int\int_B (x^2+y^2)\, dA\right).\] Let us calculate the first integral over \(A\): \[\iint_A (x^2+y^2)\,d A=\int_0^1\int_x^1(x^2+y^2)\,dydx=\int_0^1\left(x^2(1-x)+\frac13(1-x^3)\right)\,d x= \frac 13. \] On the other hand, the integral over \(B\) can be calculated by turning it into polar coordinates: \[\iint_B (x^2+y^2)\, dA=\int_{-\pi/2}^{\pi/4} \int_0^{\sqrt 2}\rho^3\,d\rho d\theta=\frac{3\pi}4\cdot \frac{4}4=\frac{3\pi}{4}.\] Therefore \(\iint_D (x^2+y^2)\,dA = 2\cdot \left(\frac13+\frac{3\pi}4\right)=\frac{4+9\pi}{6}\).

The solid that maximizes the integral is the one on which the function \(f(x,y,z)=9-x^2-y^2-4z^2\) is non-negative. Therefore the required solid is: \(x^2+y^2+9z^2\leq 9\). In order to evaluate the integral we use the substitution \(x=3\rho\cos\theta\sin\phi\), \( y=3\rho\sin\theta\sin\phi\), \(z=\rho\cos\phi\). The bounds are \(0\leq \theta\lneq 2\pi\), \(0\leq \phi\leq \pi\), \(0\leq \rho\leq 1\). The Jacobian is \(J=9\rho^2\sin\phi\) and the integral becomes: \begin{eqnarray*} I(S)&=&9\int_0^{2\pi}\int_0^1\int_0^{\pi}(9-9\rho^2)\rho^2\sin\phi\,d\phi d\rho d\theta\newline&=& 9\int_0^{2\pi}\int_0^1(9\rho^2-9\rho^4)\cdot2\,d\rho d\theta\newline&=&81\cdot 2\pi\cdot 2\cdot\left(\frac13-\frac15\right)\newline &=&\frac{216\pi}5 . \end{eqnarray*}

The suggested transformation is linear, hence the parallelogram \(R\) will be mapped to a parallelogram in the \(uv\) plane. We will show that the vertices of \(R\) are mapped to the vertices of \(\left[0,\frac{\pi}2\right]\times \left[0,\frac{\pi}2\right]\). Let us denote \(T(x,y)=(x-y,x+2y)\). Then we have \(T(0,0)=(0,0)\), \(T\left(\frac{\pi}3,-\frac{\pi}6\right)=\left(\frac{\pi}2, 0\right)\), \(T\left(\frac{\pi}2,0\right)=\left(\frac{\pi}2,\frac{\pi}2\right)\), and \(T\left(\frac{\pi}6,\frac{\pi}6\right)= \left(0, \frac{\pi}6 \right)\).

We use the substitution \(u=x-y\), \(v=x+2y\). Then \(R\) gets mapped to \(\left[0,\frac{\pi}2\right]\times \left[0,\frac{\pi}2\right]\). We first solve for \(x\) and \(y\) to obtain \(x=\frac{2u+v}3\), \(y=\frac{-u+v}3\). The jacobian is equal to \[J=\left|\det \left|\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\newline \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|\right|= \left|\det \left|\begin{array}{cc} \frac23 & \frac13\newline -\frac13& \frac13 \end{array}\right|\right|=\frac13.\] Hence we have \begin{eqnarray*} \iint_R(x-y)^3\cos(x+2y)\,dxdy&=&\int_0^{\frac{\pi}2}\int_0^{\frac{\pi}2} u^3\cos v \cdot \frac13\,dudv \\ &=&\frac13\int_0^{\frac{\pi}2}\left(\left. \cos v\cdot \frac{u^4}4\right|_{u=0}^{u=\frac{\pi}2}\right)\,dv\\&=& \frac{\pi^4}{3\cdot 4\cdot 2^4}\cdot\int_0^{\frac{\pi}2}\cos v\,dv = \frac{\pi^4}{192}. \end{eqnarray*}

Draw the picture! You may wish to draw this picture a couple of times since it is unlikely to draw the correct picture from the first attempt. After that we can change the variables to spherical coordinates: \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), \(z=\rho\cos\phi\); \(0\leq\rho\leq 1\), \(\pi/3\leq \theta\leq 2\pi/3\), \(0\leq \phi\leq \pi/2\). The integral now becomes: \begin{eqnarray*}&& \int_{\pi/3}^{2\pi/3}\int_0^{\pi/2}\int_0^1 e^{\rho\sin\phi}\rho^2\sin\phi\cdot\frac{\rho\cos\phi}{\rho^3}\,d\rho d\phi d\theta \\&=& \int_{\pi/3}^{2\pi/3}\int_0^{\pi/2}\cos\phi\left(\int_0^1e^{\rho\sin\phi}\sin\phi \,d\rho\right)d\phi d\theta\\&=& \int_{\pi/3}^{2\pi/3}\int_0^{\pi/2}\cos\phi \left(e^{\sin\phi}-1\right)\,d\phi d\theta\\&=& \int_{\pi/3}^{2\pi/3} (e-1-1)\,d\theta=\frac{\pi(e-2)}3. \end{eqnarray*}