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

question 98

Multiple Choice

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

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