Gustaf Rydberg (1835-1933) was a Swedish landscape painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Double Integrals.
Let \(\Omega\) be a closed bounded region in the \(xy\)-plane. We assume that \(\Omega\) is a basic region. A basic region is a region whose boundary consists of a finite number of continuous arcs \(y=\phi (x)\) and \(x=\psi (y)\). Let \(f\) be a nonnegative and continuous function on \(\Omega\). The double integral
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy.\]
gives the volume of the solid whose upper boundary is the surface \(z=f(x,y)\) and whose lower boundary is the region \(\Omega\); that is, the volume bounded by \(z=f(x,y)\) and \(\Omega\).
Theorem. Let \(\Omega\) be a basic region. The functions \(f\) and \(g\) are assumed to be continuous on \(\Omega\). Then, we have the following properties.
(i) We have the linearity given by
\[ \int\!\!\!\!\!\int_{\Omega}[\alpha f(x,y)+\beta g(x,y)]dxdy=\alpha \int\!\!\!\!\!\int_{\Omega}f(x,y)dxdy+\beta \int\!\!\!\!\!\int g(x,y)dxdy.\]
(ii) We have the order given by
\[\mbox{if \(f\leq g\) on \(\Omega\), then } \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy\leq \int\!\!\!\!\!\int_{\Omega} g(x,y)dxdy.\]
(iii) Suppose that \(\Omega\) is broken into a finite number of non-overlapping basic regions \(\Omega_{1},\Omega_{2},\cdots ,\Omega_{n}\). Then, we have
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy= \int\!\!\!\!\!\int_{\Omega_{1}} f(x,y)dxdy+\cdots + \int\!\!\!\!\!\int_{\Omega_{2}} f(x,y)dxdy.\]
(iv) There is a point \((x_{0},y_{0})\) in \(\Omega\) satisfying
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y_dxdy=f(x_{0},y_{0})\cdot\mbox{(area of \(\Omega\))}.\]
We call \(f(x_{0},y_{0})\) the average value of \(f\) on \(\Omega\). \(\sharp\)
Theorem. (Mean-Value Theorem for Double Integrals). Let \(f\) and \(g\) be continuous on the basic region \(\Omega\). Suppose that \(g\) is nonnegative on \(\Omega\). Then, there exists a point \((x_{0},y_{0})\) in \(\Omega\) satisfying
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)g(x,y)dxdy=f(x_{0},y_{0}) \int\!\!\!\!\!\int_{\Omega}g(x,y)dxdy.\]
The double integrals can be treated as the iterated integrals.
- Type I Region. Suppose that the region \(\Omega\) can be written as \(a\leq x\leq b\) and \(\phi_{1}(x)\leq y\leq\phi_{2}(x)\). Then, we have
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy=\int_{a}^{b}\left (\int_{\phi_{1}(x)}^{\phi_{2}(x)} f(x,y)dy\right )dx.\] - Type II Region. Suppose that the region \(\Omega\) can be written as \(c\leq y\leq d\) and \(\psi_{1}(y)\leq x\leq\psi_{2}(y)\),. Then, we have
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy=\int_{c}^{d}\left (\int_{\psi_{1}(y)}^{\psi_{2}(y)} f(x,y)dx\right )dy.\]
Example. Evaluate the double integral
\[ \int\!\!\!\!\!\int_{\Omega} (xy-y^{3})dxdy\]
with \(\Omega\) consisting of \(0\leq y\leq 1\) and \(-1\leq x\leq y\).
\begin{align*}
\int\!\!\!\!\!\int_{\Omega} (xy-y^{3})dxdy & =\int_{0}^{1}\left (\int_{-1}^{y}(xy-y^{3})dx\right )dy\\
& =\int_{0}^{1}\left [\frac{1}{2}x^{2}y-xy^{3}\right ]_{-1}^{y}dy\\
& =\int_{0}^{1}\left (-\frac{1}{2}y^{3}-y^{4}-\frac{1}{2}y\right )dy\\
& =-\frac{23}{40}.
\end{align*}
We can also rewrite the region \(\Omega\) as \(-1\leq x\leq 1\) and \(\phi (x)\leq y\leq 1\), where
\[\phi (x)=\left\{\begin{array}{ll}
0, & -1\leq x\leq 0\\
x, & 0\leq x\leq 1
\end{array}\right .\]
Then, we have
\begin{align*}
\int\!\!\!\!\!\int_{\Omega} (xy-y^{3})dxdy & =\int_{-1}^{1}\left [
\int_{\phi (x)}^{1}(xy-y^{3})dy\right ]\\ & =
\int_{-1}^{0}\left [\int_{\phi (x)}^{1}(xy-y^{3})dy\right ]+
\int_{0}^{1}\left [\int_{\phi (x)}^{1}(xy-y^{3})dy\right ]\\
& =\int_{-1}^{0}\int_{0}^{1} (xy-y^{3})dydx+\int_{0}^{1}\int_{x}^{1}
(xy-y^{3})dydx\\ & =\left (-\frac{1}{2}\right )+\left (-\frac{3}{40}\right )=-\frac{23}{40}
\end{align*}
We can see that if \(f(x,y)=1\), then
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy= \int\!\!\!\!\!\int_{\Omega}dxdy\]
is the area of \(\Omega\).
Example. Calculate the area of the region \(\Omega\) that lies between \(\sqrt{x}+\sqrt{y}=\sqrt{a}\) and \(x+y=a\). The region \(\Omega\) can be written as
$0\leq x\leq a$ and \((\sqrt{a}-\sqrt{x})^{2}\leq y\leq a-x\). Therefore, the area is given by
\begin{align*}
\int\!\!\!\!\!\int_{\Omega}dxdy & =\int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^{2}}^{a-x}dydx\\
& =\int_{0}^{a}(-2x+2\sqrt{ax})dx\\
& =\frac{a^{2}}{3}.
\end{align*}
We can also rewrite the region \(\Omega\) as \(0\leq y\leq a\) and \((\sqrt{a}-\sqrt{y})^{2}\leq x\leq a-y\). Then, we have
\[ \int\!\!\!\!\!\int_{\Omega} dxdy=\int_{0}^{a}\int_{(\sqrt{a}-\sqrt{y})^{2}}^{a-y}dxdy=\frac{1}{3}a^{2}.\]
When two orders of integration are possible, one order may be easy to carry out, while the other may present serious difficulties. Consider the double integral
\[ \int\!\!\!\!\!\int_{\Omega}\cos\frac{1}{2}\pi x^{2}dxdx=\int_{0}^{1}\int_{0}^{x}\cos
\frac{1}{2}\pi x^{2}dydx=\int_{0}^{1}\int_{y}^{1}\cos\frac{1}{2}\pi x^{2}dxdy.\]
The first double integral is easy to calculate given by
\begin{align*} \int_{0}^{1}\int_{0}^{x}\cos\frac{1}{2}\pi x^{2}dydx & =\int_{0}^{1}\left [
y\cos\frac{1}{2}\pi x^{2}\right ]_{0}^{x}dx\\ & =\int_{0}^{1}x\cos\frac{1}{2}
\pi x^{2}dx\\ & =\left [\frac{1}{\pi}\sin\frac{1}{2}\pi x^{2}\right ]_{0}^{1}=\frac{1}{\pi}.\end{align*}
However, the second integral is not simple, since \(\cos\frac{1}{2}\pi x^{2}\) does not have an elementary antiderivative. One possible way to handle the integral \(\int_{y}^{1}\cos\frac{1}{2}\pi x^{2}dx\) is to expand \(\cos\frac{1}{2}\pi x^{2}\) as a series in \(x\) and then integrate term by term. \(\sharp\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Change Variables in Multiple Integration.
Let \(\Gamma\) be a basic region in a plane that we are calling the \(uv\)-plane. Suppose that the function \(x=x(u,v)\) and \(y=y(u,v)\) are continuously differentiable on \(\Gamma\). As \((u,v)\) ranges over \(\Gamma\), the point \((x,y)=(x(u,v),y(u,v))\) generates a region \(\Omega\) in the \(xy\)-plane. Suppose that \(f(x,y)\) is a continuous function on \(\Omega\). Then, the mapping \((u,v)\rightarrow (x,y)\) is one-to-one on the interior of \(\Gamma\) and the Jacobian \(J\) given by
\[J(u,v)=\left |\begin{array}{cc}
\partial x/\partial u & \partial y/\partial u\\
\partial x/\partial v & \partial y/\partial v
\end{array}\right |=\frac{\partial x}
{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}
\frac{\partial y}{\partial u}\]
is never zero on the interior. Therefore, we have
\[ \int\!\!\!\!\!\int_{\Omega}f(x,y)dxdy= \int\!\!\!\!\!\int_{\Gamma} f(x(u,v),y(u,v))|J(u,v)|dudv.\]
Example. Evaluate the double integral
\[ \int\!\!\!\!\!\int_{\Omega} (x+y)^{2}dxdy,\]
where \(\Omega\) is the parallelogram bounded by the lines \(x+y=0\), \(x+y=1\), \(2x-y=0\), and \(2x-y=3\). The boundaries suggest that we set \(u=x+y\) and \(v=2x-y\). Then, we get
\[x=x(u,v)=\frac{u+v}{3}\mbox{ and }y=y(u,v)=\frac{2u-v}{3}.\]
The region \(\Gamma\) is bounded by \(0\leq u\leq 1\) and \(0\leq v\leq 3\), and the Jacobian is given by
\[J(u,v)=\left |\begin{array}{cc}
{\displaystyle \frac{\partial}{\partial u}\left (\frac{u+v}{3}\right )} & {\displaystyle \frac{\partial}{\partial u}\left (\frac{2u-v}{3}\right )}\\
{\displaystyle \frac{\partial}{\partial v}\left (\frac{u+v}{3}\right )} & {\displaystyle \frac{\partial}{\partial v}\left (\frac{2u-v}{3}\right )}\end{array}\right |=\left |\begin{array}{cc}{\displaystyle \frac{1}{3}} & {\displaystyle \frac{2}{3}}\\
{\displaystyle \frac{1}{3}} & -{\displaystyle \frac{1}{3}}\end{array}\right |=-\frac{1}{3}.\]
Therefore, we obtain
\begin{align*} \int\!\!\!\!\!\int_{\Omega} (x+y)^{2}dxdy & = \int\!\!\!\!\!\int_{\Gamma}u^{2}|J(u,v)|dudv\\ & =\frac{1}{3}\int_{0}^{3}\int_{0}^{1} uvdudv=\frac{1}{3}.\end{align*}
Example. Evaluate the double integral
\[ \int\!\!\!\!\!\int_{\Omega} xydxdy\]
where \(\Omega\) is the first-quadrant region bounded by the curves \(x^{2}+y^{2}=4\), \(x^{2}+y^{2}=9\), \(x^{2}-y^{2}=1\), and \(x^{2}-y^{2}=4\). The boundaries suggest that we set \(u=x^{2}+y^{2}\) and \(v=x^{2}-y^{2}\). Then, we get
\[x=\sqrt{\frac{u+v}{2}}\mbox{ and }y=\sqrt{\frac{u-v}{2}}.\]
Th region \(\Gamma\) is bounded by \(4\leq u\leq 9\) and \(1\leq v\leq 4\), and the Jacobian is
\[J(u,v)=\left |\begin{array}{ll}
{\displaystyle \frac{\partial}{\partial u}\left (\frac{\sqrt{u+v}}{2}\right )} & {\displaystyle \frac{\partial}{\partial u}\left (\frac{\sqrt{u-v}}{2}\right )}\\
{\displaystyle \frac{\partial}{\partial v}\left (\frac{\sqrt{u+v}}{2}\right )} & {\displaystyle \frac{\partial}{\partial v}\left (\frac{\sqrt{u-v}}{2}\right )}
\end{array}\right |=-\frac{1}{4\sqrt{u^{2}-v^{2}}}.\]
Therefore, we obtain
\begin{align*} \int\!\!\!\!\!\int_{\Omega} xydxdy & = \int\!\!\!\!\!\int_{\Gamma}\left (\frac{\sqrt{u+v}}{2}\right )\left (\frac{\sqrt{u-v}}{2}\right )\left (\frac{1}
{4\sqrt{u^{2}-v^{2}}}\right )dudv\\ & =\frac{1}{8}\int_{1}^{4}\int_{4}^{9}dudv=\frac{15}{8}.\end{align*}
Now, we consider the changing variables from the rectangular coordinates \((x,y)\) to polar coordinates \((r,\theta )\). Let \(x=r\cos\theta\) and \(y=r\sin\theta\). The Jacobian is
\[J(r,\theta )=\left |\begin{array}{ll}
{\displaystyle \frac{\partial}{\partial r}(r\cos\theta )} &
{\displaystyle \frac{\partial}{\partial r}(r\sin\theta )}\\
{\displaystyle \frac{\partial}{\partial \theta}(r\cos\theta )} &
{\displaystyle \frac{\partial}{\partial \theta}(r\sin\theta )}
\end{array}\right |=r(\cos^{2}\theta +\sin^{2}\theta )=r.\]
Therefore, we obtain
\[ \int\!\!\!\!\!\int_{\Omega} f(x,y)dxdy= \int\!\!\!\!\!\int_{\Gamma} f(r\cos\theta ,r\sin\theta )rdrd\theta .\]
Example. Calculate the volume of the solid bounded above by the cone \(z=2-\sqrt{x^{2}+y^{2}}\) and bounded below by the disc \(\Omega\): $latex (x-1)^{2}+
y^{2}\leq 1$. The volume is
\[V= \int\!\!\!\!\!\int_{\Omega}(2-\sqrt{x^{2}+y^{2}})dxdy=2 \int\!\!\!\!\!\int_{\Omega}dxdy- \int\!\!\!\!\!\int_{\Omega}\sqrt{x^{2}+y^{2}}dxdy.\]
The first integral is \(2*(\)area of \(\Omega )=2\pi\). We evaluate the second integral by changing to polar coordinates. The equation \((x-1)^{2}+y^{2}=1\) simplifies to \(x^{2}+y^{2}=2x\). In polar coordinates, this becomes \(r^{2}=2r\cos\theta\), which simplifies to \(r=2\cos\theta\). The disc \(\Omega\) is the set of all points with polar coordinates in the set \(\Gamma\): \(-\frac{1}{2}\pi\leq\theta\leq\frac{1}{2}\pi\) and \(0\leq r\leq 2\cos\theta\). Therefore, we obtain
\begin{align*} \int\!\!\!\!\!\int_{\Omega}\sqrt{x^{2}+y^{2}}dxdy & = \int\!\!\!\!\!\int_{\Gamma} r^{2}drd\theta \\ & =\int_{-\pi /2}^{\pi /2}\int_{0}^{2\cos\theta} r^{2}drd\theta =\frac{32}{9}.\end{align*}
Then, we have \(V=2\pi-\frac{32}{9}\). \(\sharp\)
Example. The function \(f(x)=e^{-x^{2}}\) has no elementary antiderivative. Nevertheless, using polar coordinates, we are going to obtain
\[\int_{-\infty}^{\infty} e^{-x^{2}}dx=\sqrt{\pi}.\]
We consider the following integrals
\begin{align*}
&\left (\int_{-\infty}^{\infty} e^{-x^{2}}dx\right )^{2}=\left (\int_{-\infty}^{\infty} e^{-x^{2}}dx\right )
\left (\int_{-\infty}^{\infty} e^{-x^{2}}dx\right )\\ & =\left (\int_{-\infty}^{\infty} e^{-x^{2}}dx\right )
\left (\int_{-\infty}^{\infty} e^{-y^{2}}dx\right )=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})}dxdy.
\end{align*}
Let \(x=r\cos\theta\) and \(y=r\sin\theta\). We have
\begin{align*} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})}dxdy & =
\int_{0}^{2\pi}\int_{0}^{\infty} re^{-r^{2}}drd\theta\\ & =\int_{0}^{2\pi}
\left [-\frac{1}{2}e^{-r^{2}}\right ]_{0}^{\infty}d\theta\\ & =\frac{1}{2}\int_{0}^{2\pi}d\theta=\pi .\end{align*}
The proof is complete. \(\sharp\)
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Triple Integrals.
Let \(T\) be the set of all points \((x,y,z)\) with \(a_{1}\leq x\leq a_{2}\), \(\phi_{1}(x)\leq y\leq\phi_{2}(x)\), and \(\psi_{1}(x,y)\leq z\leq\psi_{2}(x,y)\). The triple integral over \(T\) can be expressed as
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T} f(x,y,z)dxdydz=\int_{a_{1}}^{a_{2}}\int_{\phi_{1}(x)}^{\phi_{2}(x)}\int_{\psi_{1}(x,y)}^{\psi_{2}(x,y)}f(x,y,z)dzdydx.\]
There is nothing special about this order of integration. Other orders of integration are possible and in some cases more convenient. If the set \(T\) is the set of all point \((x,y,z)\) with \(a_{1}\leq z\leq a_{2}\), \(\phi_{1}(z)\leq x\leq \phi_{1}(z)\) and \(\psi_{1}(x,z)\leq y\leq\psi_{2}(x,z)\), then
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T} f(x,y,z)dxdydz=\int_{a_{1}}^{a_{2}}\int_{\phi_{1}(z)}^{\phi_{2}(z)}\int_{\psi_{1}(x,z)}^{\psi_{2}(x,z)}
f(x,y,z)dydxdz.\]
We can also see that the volume of \(T\) is, when \(f(x,y,z)=1\),
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T}dxdydz.\]
Example. Evaluate the triple integral
\begin{align*}
\int_{0}^{2}\int_{0}^{x}\int_{0}^{4-x^{2}}xyzdzdydx & =\int_{0}^{2}\int_{0}^{x}\left (\left [\frac{1}{2}xyz^{2}\right ]_{0}^{4-x^{2}}\right )dydx\\
& =\frac{1}{2}\int_{0}^{2}\int_{0}^{x}x(4-x^{2})^{2}ydydx\\
& =\frac{1}{2}\int_{0}^{2}\left (\left [\frac{1}{2}x(4-x^{2})^{2}y^{2}\right ]_{0}^{x}\right )dx\\
& =\frac{1}{4}\int_{0}^{2}x^{3}(4-x^{2})^{2}dx=\frac{8}{3}.
\end{align*}
Example. Use the triple integration to find the volume of the solid \(T\) bounded above by the parabolic cylinder \(z=4-y^{2}\) and bounded below by the elliptic paraboloid \(z=x^{2}+3y^{2}\). Solving the two equations simultaneously, we have \(x^{2}+4y^{2}=4\). This tells us that the two surfaces intersect in a space curve that lies along the elliptic cylinder \(x^{2}+4y^{2}=4\). The projection of this intersection onto the \(xy\)-plane is the ellipse \(x^{2}+4y^{2}=4\). The projection of \(T\) onto the \(xy\)-plane is the region
\[\Omega_{xy}: -2\leq x\leq 2\mbox{ and }-\frac{1}{2}\sqrt{4-x^{2}}\leq y\leq\frac{1}{2}\sqrt{4-x^{2}}.\]
The solid \(T\) is the set of all points \((x,y,z)\) with \(-2\leq x\leq 2\), \(-\frac{1}{2}\sqrt{4-x^{2}}\leq y\leq\frac{1}{2}\sqrt{4-x^{2}}\), and \(x^{2}+3y^{2}\leq x\leq 4-y^{2}\). Therefore, the volume is given by
\[\int_{-2}^{2}\int_{-\sqrt{4-x^{2}}/2}^{\sqrt{4-x^{2}}/2}\int_{x^{2}+3y^{2}}^{4-y^{2}}dzdydx=4\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}/2}
\int_{x^{2}+3y^{2}}^{4-y^{2}}dzdydx=4\pi.\]
When changing variables in a triple integral, we make three coordinate changes: \(x=x(u,v,w)\), \(y=y(u,v,w)\) and \(z=z(u,v,w)\). If these functions carry a basic solid \(\Gamma\) onto a solid \(T\), then, under conditions analogous to the two-dimensional case, we have
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T} f(x,y,z)dxdydz= \int\!\!\!\!\!\int\!\!\!\!\!\int_{\Gamma} f(x(u,v,w),y(u,v,w),z(u,v,w))|J(u,v,w)|dudvdw,\]
where the Jacobian \(J(u,v,w)\) is given by
\[J(u,v,w)=\left |\begin{array}{ccc}
\partial x/\partial u & \partial y/\partial u & \partial z/\partial u\\
\partial x/\partial v & \partial y/\partial v & \partial z/\partial v\\
\partial x/\partial w & \partial y/\partial w & \partial z/\partial w
\end{array}\right |.\]
Now, we consider the cylindrical coordinates. The rectangular coordinates \((x,y,z)\) are related to cylindrical coordinates \((r,\theta ,z)\) by the equations \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\). Conversely, cylindrical coordinates \((r,\theta ,z)\) are related to rectangular coordinates by \(r=\sqrt{x^{2}+y^{2}}\), \(\tan\theta =y/x\) and \(z=z\). Suppose that \(T\) is some basic solid in \(xyz\)-space. If \(T\) is the set of all \((x,y,z)\) with the cylindrical coordinates in some basic solid \(S\) in \(r\theta z\)-space, then
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T} f(x,y,z)dxdydz= \int\!\!\!\!\!\int\!\!\!\!\!\int_{S} f(r\cos\theta ,r\sin\theta ,z)rdrd\theta dz\]
since the Jacobian is \(J(r,\theta ,z)=r\).
Example. Evaluate the triple integral
\[\int_{-2}^{2}\int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}\int_{0}^{4-x^{2}-y^{2}}(x^{2}+y^{2})dzdydx.\]
The solid \(T\) is given by \(-2\leq x\leq 2\), \(-\sqrt{4-x^{2}}\leq y\leq\sqrt{4-x^{2}}\), and \(0\leq z\leq 4-x^{2}-y^{2}\). This solid is bounded above by the paraboloid of revolution \(z=4-x^{2}-y^{2}\) and below by the \(xy\)-plane. Since the solid is symmetric about the \(z\)-axis, the integral has a simple representation in cylindrical coordinates: \(S\): \(0\leq r\leq 2\), \(0\leq\theta\leq 2\pi\), and \(0\leq z\leq r^{2}\). Therefore, we have
\begin{align*}
\int\!\!\!\!\!\int\!\!\!\!\!\int_{T} (x^{2}+y^{2})dxdydz & = \int\!\!\!\!\!\int\!\!\!\!\!\int_{S} r^{2}rdrd\theta dz\\
& =\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{4-r^{2}}r^{3}dzdrd\theta\\
& =\int_{0}^{2\pi}\int_{0}^{2} (4r^{3}-r^{5})drd\theta\\
& =\int_{0}^{2\pi}\left [r^{4}-\frac{1}{6}r^{6}\right ]_{0}^{2}d\theta =\frac{32\pi}{3}.
\end{align*}
Now, we consider the spherical coordinates \((\rho ,\theta ,\phi )\). The rectangular coordinates \((x,y,z)\) are related to spherical coordinates \((\rho ,\theta ,\phi )\) by the equations: \(x=\rho\sin\phi\cos\theta\), \(y=\rho\sin\phi\sin\theta\), and \(z=\rho\cos\phi\). Conversely, the spherical coordinates \((\rho ,\theta ,\phi )\) are related to rectangular coordinates \((x,y,z)\) by the equations
\[\rho=\sqrt{x^{2}+y^{2}+z^{2}},\tan\theta =\frac{y}{x},\cos\phi =\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}.\]
Suppose that \(T\) is basic solid in \(xyz\)-space with spherical coordinates in some basic solid \(S\) of \(\rho\theta\phi\)-space. Then, we have
\[ \int\!\!\!\!\!\int\!\!\!\!\!\int_{T} f(x,y,z)dxdydz= \int\!\!\!\!\!\int\!\!\!\!\!\int_{S} f(\rho\sin\phi\cos\theta ,
\rho\sin\phi\sin\theta ,\rho\cos\phi )\rho^{2}\sin\phi d\rho d\theta d\phi \]
since the Jacobian \(J(\rho ,\theta ,\phi )\) is \(\rho^{2}\sin\phi\).
Example. Find the volume of the solid \(T\) enclosed by the surface \((x^{2}+y^{2}+z^{2})^{2}=2z(x^{2}+y^{2})\). We have \(\rho^{4}=2\rho\cos\phi\rho^{2}\sin^{2}\phi\). In spherical coordinates the bounding surface takes the form \(\rho =2\sin^{2}\phi\cos\phi\). This equation places no restriction on \(\theta\); thus \(\theta\) can range from \(0\) to \(2\pi\). Since \(\rho\) remains nonnegative, \(\phi\) can range onl from \(0\) to \(\frac{1}{2}\pi\). Thus the solid \(T\) is the set of all \((x,y,z)\) with spherical coordinates \((\rho ,\theta ,\phi )\) in the set \(S\): \(0\leq\theta \leq 2\pi\), \(0\leq\phi\leq\frac{1}{2}\pi\) (since \(\rho=2\sin^{2}\phi\cos\phi \geq 0\)), and \(0\leq\rho\leq 2\sin^{2}\phi\cos\phi\). Therefore, we obtain
\begin{align*}
\int\!\!\!\!\!\int\!\!\!\!\!\int_{T}dxdydz & = \int\!\!\!\!\!\int\!\!\!\!\!\int_{S}\rho^{2}\sin\phi d\rhod\theta d\phi\\
& =\int_{0}^{2\pi}\int_{0}^{\pi /2}\int_{0}^{2\sin^{2}\phi\cos\phi}\rho^{2}\sin\phi d\rho d\theta d\phi\\
& =\int_{0}^{2\pi}\int_{0}^{\pi/2}\frac{8}{3}\sin^{7}\phi\cos^{3}\phid\phi d\theta\\
& =\frac{8}{3}\left (\int_{0}^{2\pi}d\theta\right )\left (\int_{0}^{\pi /2} (\sin^{7}\phi\cos\phi -\sin^{9}\phi\cos\phi )d\phi\right )\\
& =\frac{8}{3}\cdot 2\pi\cdot\frac{1}{40}=\frac{2}{15}\pi.
\end{align*}
