Set up a double integral that gives the area of the surface of the graph of f over the region R. $f ( x , y ) = e ^ { 9 x y }$
$R = \{ ( x , y ) : 0 \leq x \leq 6,0 \leq y \leq 2 \}$
A) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 81 e ^ { 18 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
B)$\int_{0}^{2} \int_{0}^{6} \sqrt{1+81 e^{18 x y}\left(x^{2}+y^{2}\right)} d y d x$
C)$\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 81 e ^ { 9 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
D) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 9 e ^ { 9 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
E) $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 2 } \sqrt { 1 + 81 e ^ { 18 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$

Question

Set up a double integral that gives the area of the surface of the graph of f over the region R. $f ( x , y ) = e ^ { 9 x y }$
$R = \{ ( x , y ) : 0 \leq x \leq 6,0 \leq y \leq 2 \}$
A) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 81 e ^ { 18 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
B)$\int_{0}^{2} \int_{0}^{6} \sqrt{1+81 e^{18 x y}\left(x^{2}+y^{2}\right)} d y d x$ 
C)$\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 81 e ^ { 9 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
D) $\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 6 } \sqrt { 1 + 9 e ^ { 9 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$
E) $\int _ { 0 } ^ { 6 } \int _ { 0 } ^ { 2 } \sqrt { 1 + 81 e ^ { 18 x y } \left( x ^ { 2 } + y ^ { 2 } \right) } d x d y$

Quizplus · Accepted Answer

The correct answer is A because the integrand is the correct formula for the surface area of a graph, and the limits of integration are correct for the given region R.

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