Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.
-$x = \int _ { 0 } ^ { y } \cos t d t , 0 \leq y \leq \pi / 3 ; y \text {-axis }$
A) $\pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \sin t d t \right) \cos y d y \quad$
B) $- 2 \pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \sin t d t \right) \cos y d y$
C) $- 2 \pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \cos t d t \right) \sin y d y$

D) $\pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \cos t d t \right) \sin y d y$

Question

Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.
-$x = \int _ { 0 } ^ { y } \cos t d t , 0 \leq y \leq \pi / 3 ; y \text {-axis }$
A) $\pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \sin t d t \right) \cos y d y \quad$ 
B) $- 2 \pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \sin t d t \right) \cos y d y$ 
C) $- 2 \pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \cos t d t \right) \sin y d y$

D) $\pi \int _ { 0 } ^ { \pi / 3 } \left( \int _ { 0 } ^ { y } \cos t d t \right) \sin y d y$

Quizplus · Accepted Answer

The formula for the surface area of a curve $ x = f(y) $ revolved around the y-axis is $ 2\pi \int f(y) \sqrt{1 + (f'(y))^2} \, dy $. Here, $ f(y) = \int_0^y \cos t \, dt $ and $ f'(y) = \cos y $. The integral becomes $-2\pi \int_0^{\pi/3} \left( \int_0^y \cos t \, dt \right) \sin y \, dy$.

Set Up an Integral for the Area of the Surface x=∫0ycos⁡tdt,0≤y≤π/3;y-axis x = \int _ { 0 } ^ { y } \cos t d t , 0 \leq y \leq \pi / 3 ; y \text {-axis }x=∫0y​costdt,0≤y≤π/3;y-axis

Set Up an Integral for the Area of the Surface $x = \int _ { 0 } ^ { y } \cos t d t , 0 \leq y \leq \pi / 3 ; y \text {-axis }$