Set Up and Evaluate the Integral That Gives the Volume $y = 8$

question 52

Multiple Choice

Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded by $y = 8$ and $y = 16 - \frac { x ^ { 2 } } { 16 }$ about the $x$ -axis.

• A)\(V = \pi \int _ { - 16 } ^ { 16 } \left( \left( 16 - \frac { x ^ { 2 } } { 16 } \right) ^ { 2 } - 64 \right) d x = \frac { 7168 } { 15 } \sqrt { 2 } \pi\)
• B)\(V = \pi \int _ { - 8 \sqrt { 2 } } ^ { 8 \sqrt { 2 } } \left( \left( 16 - \frac { x ^ { 2 } } { 16 } \right) ^ { 2 } - 64 \right) d x = \frac { 14336 } { 15 } \sqrt { 2 } \pi\)
• C)\(V = \pi \int _ { - 16 } ^ { 16 } \left( \left( 16 - \frac { x ^ { 2 } } { 16 } \right) ^ { 2 } - 64 \right) d x = \frac { 28672 } { 15 } \sqrt { 2 } \pi\)
• D)\(\left. V = \pi \int _ { - 8 \sqrt { 2 } } ^ { 8 \sqrt { 2 } } \left( 16 - \frac { x ^ { 2 } } { 16 } \right) ^ { 2 } - 64 \right) d x = \frac { 28672 } { 15 } \sqrt { 2 } \pi\)
• E)\(V = \pi \int _ { - 16 } ^ { 16 } \left( \left( 16 - \frac { x ^ { 2 } } { 16 } \right) ^ { 2 } - 64 \right) d x = \frac { 14336 } { 15 } \sqrt { 2 } \pi\)

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