Now that we understand what double integrals are, we can use them to compute areas of regions and volumes of solids of revolution. The following video shows how.
The area of a region $R$ is just $\iint_R 1 \,dA$. Rewriting that as an iterated integral gives us the formulas for area that we derived a long time ago. If $R$ is a Type I region, bounded by $y=g(x)$, $y=h(x)$, $x=a$ and $x=b$, then the area of $R$ is \begin{eqnarray*} \iint_R 1 \, dA &=& \int_a^b \int_{g(x)}^{h(x)} 1\, dy\,dx\\ &=& \int_a^b \Bigl(h(x)-g(x)\Bigr) dx. \end{eqnarray*} This is our familiar formula for the area between two curves.
If $R$ is a Type II region, bounded by $x=g(y)$, $x=h(y)$, $y=c$ and $y=d$, then the area of $R$ is \begin{eqnarray*} \iint_R 1 \, dA &=& \int_c^d \int_{g(y)}^{h(y)} 1\, dx\,dy\\ &=& \int_c^d \Bigl(h(y) - g(y)\Bigr)\, dx. \end{eqnarray*}
Now suppose that $R$ lies above the $x$-axis. If we rotate the region $R$ around the $x$ axis, then the volume of the solid of revolution is $\displaystyle{\iint_R 2\pi y \,dA}$. If $R$ is a Type I region and we integrate first over $y$ and then over $x$, we get \begin{eqnarray*} \iint_R 2 \pi y dA & = & \int_a^b \int_{g(x)}^{h(x)} 2 \pi y \,dy\, dx \\ & = & \int_a^b \left . \pi y^2 \right |_{g(x)}^{h(x)} \,dx \\ & = & \int_a^b \pi \Bigl(h(x)^2 - g(x)^2\Bigr) dx. \end{eqnarray*} This is our familiar formula for volumes by washers. On the other hand, if $R$ is a Type II region and we integrate first over $x$, we get \begin{eqnarray*} \iint_R 2 \pi y dA & = & \int_c^d \int_{g(y)}^{h(y)} 2 \pi y \,dx\, dy \\ & = & \int_a^b 2 \pi y \Bigl(h(y) - g(y)\Bigr)\, dy. \end{eqnarray*} This is the formula for volume by cylindrical shells.
Likewise, if $R$ lies to the right of the $y$ axis and we rotate around the $y$ axis, then the volume the solid of revolution is $\displaystyle{\iint_R 2\pi x\, dA}$. Integrating $dx \, dy$ gives volume by washers and integrating $dy\,dx$ gives volume by cylindrical shells.