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, and the main formulas are
repeated below the video for convenience.
Areas: 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
$$\iint_R 1 \, dA = \int_a^b \int_{g(x)}^{h(x)} 1\, dy\,dx
= \int_a^b [h(x)-g(x)] dx.
$$
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
$$ \iint_R 1 \, dA = \int_c^d \int_{g(y)}^{h(y)} 1\, dx\,dy
= \int_c^d [h(y) - g(y)] dx.
$$
Volumes of solids of revolution:
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
$$
\iint_R 2 \pi y dA = \int_a^b \int_{g(x)}^{h(x)} 2 \pi y \,dy\, dx
= \int_a^b \pi y^2 \Big |_{g(x)}^{h(x)} dx
= \int_a^b \pi [h(x)^2 - g(x)^2] dx.
$$
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
$$\iint_R 2 \pi y dA = \int_c^d \int_{g(y)}^{h(y)} 2 \pi y \,dx\, dy
= \int_a^b 2 \pi y [h(y) - g(y)] dy.
$$
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.