An auxiliary fuel tank for a helicopter is shaped like the surface generated by revolving the curve $y = 1 - \frac { x ^ { 2 } } { 16 }$, $= 4$ $\leq x \leq 4$, about the $x$-axis (dimensions are in feet). If a cubic foot holds $7.481$ gallons and the helicopter gets 3 miles to the gallon, how many additional miles will the helicopter be able to fly once the tank is installed (to the nearest mile)?
A) 75
B) 150
C) 301
D) 100

Question

Quizplus · Accepted Answer

We need to find the volume of the auxiliary fuel tank, which can be found using the formula for the volume of a solid of revolution: $V=\int_a^b \pi (f(x))^2\, dx$, where $f(x)$ is the given curve and we are revolving around the $x$-axis. In this case, $a=-4$, $b=4$, and $f(x)=1-\frac{x^2}{16}$, so we have \begin{align*}
V&=\int_{-4}^4 \pi \left(1-\frac{x^2}{16}ight)^2\, dx \
&=\pi \int_{-4}^4 \left(1-\frac{x^2}{8}+\frac{x^4}{256}ight)\, dx \
&=\pi\left[x-\frac{x^3}{24}+\frac{x^5}{1280}ight]_{-4}^4 \
&=128\pi.
\end{align*} We then convert this volume to gallons using the given conversion factor of $1	ext{ ft}^3 = 7.481	ext{ gal}$: \begin{align*}
128\pi	ext{ ft}^3 \cdot 7.481\frac{	ext{gal}}{	ext{ft}^3} &= 960.44\pi	ext{ gal}.
\end{align*} Finally, we can find the number of additional miles the helicopter can fly by multiplying the volume in gallons by the fuel efficiency: \begin{align*}
960.44\pi	ext{ gal} \cdot 3\frac{	ext{miles}}{	ext{gallon}} &= 2881.32\pi	ext{ miles}.
\end{align*} To the nearest mile, this is $\boxed{	extbf{(C) }301}$.

An Auxiliary Fuel Tank for a Helicopter Is Shaped Like y=1−x216y = 1 - \frac { x ^ { 2 } } { 16 }y=1−16x2​

An Auxiliary Fuel Tank for a Helicopter Is Shaped Like $y = 1 - \frac { x ^ { 2 } } { 16 }$