Section 15.9 : Surface Area
2. Determine the surface area of the portion of \(z = 13 - 4{x^2} - 4{y^2}\) that is above \(z = 1\) with \(x \le 0\) and \(y \le 0\).
Show All Steps Hide All Steps
Start SolutionOkay, 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.
The surface we are after is the portion of the elliptic paraboloid (the orange surface in the sketch) that is above \(z = 1\) and in the 3rd quadrant of the \(xy\)-plane. The bluish “walls” are simply there to provide a frame of reference to help visualize the surface and are not actually part of the surface we are interested in.
Show Step 2The integral for the surface area is,
\[S = \iint\limits_{D}{{\sqrt {{{\left[ { - 8x} \right]}^2} + {{\left[ { - 8y} \right]}^2} + 1} \,dA}} = \iint\limits_{D}{{\sqrt {64{x^2} + 64{y^2} + 1} \,dA}}\] Show Step 3Now, as implied in the last step \(D\) must be in the \(xy\)-plane and it should be (hopefully) pretty obvious that it will be a circular region. If you look at the surface from directly above you will see the following region.
To determine equation of the circle all we need to do is set the equation of the elliptic paraboloid equal to \(z = 1\) to get,
\[1 = 13 - 4{x^2} - 4{y^2}\hspace{0.25in} \to \hspace{0.25in}4{x^2} + 4{y^2} = 12\hspace{0.25in} \to \hspace{0.25in}{x^2} + {y^2} = 3 \]So, it is the portion of the circle of radius \(\sqrt 3 \) in the 3rd quadrant.
At this point it should also be clear that we’ll need to evaluate the integral in terms of polar coordinates. In terms of polar coordinates the limits for \(D\) are,
\[\begin{array} \displaystyle \pi \le \theta \le \frac{{3\pi }}{2}\\ 0 \le r \le \sqrt 3 \end{array}\] Show Step 4Now, let’s convert the integral into polar coordinates.
\[S = \iint\limits_{D}{{\sqrt {64{x^2} + 64{y^2} + 1} \,dA}} = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{0}^{{\sqrt 3 }}{{r\sqrt {64{r^2} + 1} \,dr}}\,d\theta }}\]Don’t forget to pick up the extra \(r\) from converting the \(dA\) into polar coordinates. If you need a refresher on converting integrals to polar coordinates then you should go back and work some problems from that section.
Show Step 5Okay, all we need to do then is evaluate the integral.
\[\begin{align*}S & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{0}^{{\sqrt 3 }}{{r\sqrt {64{r^2} + 1} \,dr}}\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\left. {\frac{1}{{192}}{{\left( {64{r^2} + 1} \right)}^{\frac{3}{2}}}} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\frac{1}{{192}}\left( {{{193}^{\frac{3}{2}}} - 1} \right)\,d\theta }}\\ & = \left. {\frac{1}{{192}}\left( {{{193}^{\frac{3}{2}}} - 1} \right)\theta } \right|_\pi ^{\frac{3}{2}\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{\pi }{{384}}\left( {{{193}^{\frac{3}{2}}} - 1} \right) = 21.9277}}\end{align*}\]