Find the exact area of the surface obtained by rotating the given curve about the x-axis.
E)None of these

Question

Find the exact area of the surface obtained by rotating the given curve about the x-axis. 
E)None of these

Quizplus · Accepted Answer

The given curve is a paraboloid opening downwards with equation $z=16-x^2-y^2$. To find the exact area of the surface obtained by rotating this curve about the x-axis, we use the formula:

$$A = 2\pi\int_{a}^{b} f(x) \sqrt{1+\left(\frac{df}{dx}ight)^2}\,dx$$

where $f(x) = 16-x^2$, $a=-4$, and $b=4$. We have $\frac{df}{dx} = -2x$, so

$$A = 2\pi\int_{-4}^{4} (16-x^2) \sqrt{1+4x^2}\,dx$$

This integral can be evaluated using a substitution or integration by parts, but it is quite tedious. However, we can use a trick to simplify the integral:

\begin{align*}
\sqrt{1+4x^2} &= \sqrt{(2x)^2 +1}\
&= \frac{1}{\sqrt{3}}\sqrt{(2x)^2+3}
\end{align*}

Now let $u=2x$, so $dx = \frac{1}{2}du$. Then

\begin{align*}
A &= 2\pi\int_{-4}^{4} (16-x^2) \sqrt{1+4x^2}\,dx\
&= 2\pi\int_{-8}^{8} \left(16-\frac{u^2}{4}ight) \frac{1}{\sqrt{3}}\sqrt{u^2+3}\,\frac{1}{2}\,du\
&= \frac{\pi}{\sqrt{3}}\int_{-8}^{8} u\sqrt{u^2+3}-2\sqrt{u^2+3}\,du + 32\pi\int_{-8}^{8} \frac{1}{\sqrt{3}}\,du\
&= \frac{\pi}{\sqrt{3}}\left[\frac{1}{3}(u^2+3)^{3/2}-\frac{2}{3}(u^2+3)^{1/2}ight]_{-8}^{8} + 64\pi\sqrt{3}\
&= \frac{128\pi}{\sqrt{3}} - \frac{32\pi}{\sqrt{3}} + 64\pi\sqrt{3}\
&= \boxed{48\pi\sqrt{3}}
\end{align*}

Find the Exact Area of the Surface Obtained by Rotating