Set Up a Double Integral That Gives the Area of the Surface

question 24

Multiple Choice

Set up a double integral that gives the area of the surface on the graph of $f ( x , y ) = 10 \cos \left( x ^ { 2 } + y ^ { 2 } \right)$ over the region $R = \left\{ ( x , y ) : x ^ { 2 } + y ^ { 2 } \leq \frac { \pi } { 4 } \right\}$

• A)
• \(S = \int _ { - \sqrt { \pi / 4 } } ^ { \sqrt { \pi / 4 } } \sqrt { ( \pi / 4 ) - x ^ { 2 } } \sqrt { ( \pi / 4 ) - x ^ { 2 } } \sqrt { 1 + 400 \left( x ^ { 2 } + y ^ { 2 } \right) \sin ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right) } d y d x\)
• B)
• \(S = \int _ { - \sqrt { \pi / 4 } } ^ { \sqrt { \pi / 4 } } \int _ { - \sqrt { ( \pi / 4 ) - x ^ { 2 } } } ^ { \sqrt { ( \pi / 4 ) - x ^ { 2 } } } \sqrt { 1 + 400 \left( x ^ { 2 } + y ^ { 2 } \right) \cos ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right) } d y d x\)
• C)\(\begin{aligned} S = & \int _ { - \sqrt { ( \pi / 4 ) - y ^ { 2 } } } ^ { \sqrt { ( \pi / 4 ) - x ^ { 2 } } } \int ^ { 2 } - \sqrt { ( \pi / 4 ) - x ^ { 2 } } \sqrt { 3 + 400 \left( x ^ { 2 } + y ^ { 2 } \right) \cos ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right) } d y d x \end{aligned}\)
• D)
• \(\begin{aligned} S = & \int _ { - \sqrt { ( \pi / 4 ) - y ^ { 2 } } - \sqrt { ( \pi / 4 ) - x ^ { 2 } } } ^ { \sqrt { ( \pi / 4 ) - y ^ { 2 } } } \sqrt { ( \pi / 4 ) - x ^ { 2 } } \end{aligned}\)
• E)
• \(S = \int _ { - \sqrt { ( \pi / 4 ) - y ^ { 2 } } - \sqrt { ( \pi / 4 ) - x ^ { 2 } } } ^ { \sqrt { ( \pi / 4 ) - y ^ { 2 } } \sqrt { ( \pi / 4 ) - x ^ { 2 } } } \sqrt { 1 + 400 \left( x ^ { 2 } + y ^ { 2 } \right) \cos ^ { 2 } \left( x ^ { 2 } + y ^ { 2 } \right) } d y d x\)

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