Set Up an Integral for the Area of the Surface $x y = 4,2 \leq y \leq 3 ; y$

question 42

Multiple Choice

Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis.
- $x y = 4,2 \leq y \leq 3 ; y$ -axis

• A) \(8 \pi \int _ { 2 } ^ { 3 } \frac { 1 } { y } \sqrt { 1 + 4 y ^ { - 4 } } d x\)
• B) \(8 \pi \int _ { 2 } ^ { 3 } \frac { 1 } { y } \sqrt { 1 + 16 y ^ { - 4 } } d x\)
• C) \(4 \pi \int _ { 2 } ^ { 3 } \frac { 1 } { y } \sqrt { 1 + 4 y ^ { - 4 } } \mathrm { dx }\)
• D) \(4 \pi \int _ { 2 } ^ { 3 } \frac { 1 } { y } \sqrt { 1 + 16 y ^ { - 4 } } \mathrm { dx }\)

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