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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The Answer of 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) $4 \pi \int _ { 2 } ^ { 3 } \frac { 1 } { y } \sqrt { 1 + 16 y ^ { - 4 } } \mathrm { dx }$

Set Up an Integral for the Area of the Surface xy=4,2≤y≤3;yx y = 4,2 \leq y \leq 3 ; yxy=4,2≤y≤3;y

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