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### Section 15.9 : Surface Area

4. Determine the surface area of the portion of $$y = 2{x^2} + 2{z^2} - 7$$that is inside the cylinder $${x^2} + {z^2} = 4$$.

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Okay, let’s start off with a quick sketch of the surface so we can get a feel for what we’re dealing with.  We’ve given the sketches with a set of “traditional” axes as well as a set of “box” axes to help visualize the surface.

So, we have an elliptic paraboloid centered on the $$y$$-axis. This also means that the region $$D$$ for our integral will be in the $$xz$$-plane and we’ll be needing polar coordinates for the integral.

Show Step 2

As noted above the region $$D$$ is going to be in the $$xz$$-plane and the surface is given in the form $$y = f\left( {x,z} \right)$$. The formula for surface area we gave in the notes is only for a region $$D$$ that is in the $$xy$$-plane with the surface given by $$z = f\left( {x,y} \right)$$ .

However, it shouldn’t be too difficult to see that all we need to do is modify the formula in the following manner to get one for this setup.

$S = \iint\limits_{D}{{\sqrt {{{\left[ {{f_x}} \right]}^2} + {{\left[ {{f_z}} \right]}^2} + 1} \,dA}}\hspace{0.25in}\,y = f\left( {x,z} \right)\hspace{0.25in}D\,\,{\mbox{is in the }}xz{\mbox{ - plane}}$

So, the integral for the surface area is,

$S = \iint\limits_{D}{{\sqrt {{{\left[ {4x} \right]}^2} + {{\left[ {4z} \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {16{x^2} + 16{z^2} + 1} \,dA}}$ Show Step 3

Now, as we noted in Step 1 the region $$D$$ is in the $$xz$$-plane and because we are after the portion of the elliptical paraboloid that is inside the cylinder given by $${x^2} + {z^2} = 4$$ we can see that the region $$D$$ must therefore be the disk $${x^2} + {z^2} \le 4$$.

Also as noted in Step 1 we’ll be needing polar coordinates for this integral so here are the limits for the integral in terms of polar coordinates.

$\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}$ Show Step 4

Now, let’s convert the integral into polar coordinates. However, they won’t be the “standard” polar coordinates. Because $$D$$ is in the $$xz$$-plane we’ll need to use the following “modified” polar coordinates.

\begin{align*}x & = r\cos \theta \\ z & = r\sin \theta \\ {x^2} + {z^2} & = {r^2}\end{align*}

Converting the integral to polar coordinates then gives,

$S = \iint\limits_{D}{{\sqrt {16{x^2} + 16{z^2} + 1} \,dA}} = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{r\sqrt {16{r^2} + 1} \,dr}}\,d\theta }}$

Don’t forget to pick up the extra $$r$$ from converting the $$dA$$ into polar coordinates. It is the same $$dA$$ as we use for the “standard” polar coordinates.

Show Step 5

Okay, all we need to do then is evaluate the integral.

\begin{align*}S & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{r\sqrt {16{r^2} + 1} \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\frac{1}{{48}}{{\left( {16{r^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{1}{{48}}\left( {{{65}^{\frac{3}{2}}} - 1} \right)\,d\theta }}\\ & = \left. {\frac{1}{{48}}\left( {{{65}^{\frac{3}{2}}} - 1} \right)\theta } \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{24}}\left( {{{65}^{\frac{3}{2}}} - 1} \right) = 68.4667}}\end{align*}