Find the area of the surface generated by revolving the curve about the indicated axis.
-$x = 3 \sqrt { 4 - y } , 0 \leq y \leq 15 / 4$; y-axis
A) $\frac { 125 } { 2 } \pi$
B) $\left( \frac { 125 } { 2 } - 5 \sqrt { 10 } \right) \pi$
C) $5 \pi \sqrt { 10 }$
D) $\left( \frac { 125 } { 2 } + 5 \sqrt { 10 } \right) \pi$

Question

Quizplus · Accepted Answer

The surface area of a curve revolved around the y-axis is given by the integral $2\pi \int x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$. For $x = 3 \sqrt{4 - y}$, $\frac{dx}{dy} = -\frac{3}{2\sqrt{4-y}}$. Evaluating the integral from $y = 0$ to $y = \frac{15}{4}$ gives the result $\left( \frac { 125 } { 2 } - 5 \sqrt { 10 } \right) \pi$.

Find the Area of the Surface Generated by Revolving the Curve