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

question 28

Multiple Choice

Set up a double integral that gives the area of the surface of the graph of f over the region R.
$\begin{array} { l } f ( x , y ) = x ^ { 2 } - 9 x y + 9 y ^ { 2 } \\R = \{ ( x , y ) : - 6 \leq x \leq 6 , - 8 \leq y \leq 8 \}\end{array}$

• A) \(\int _ { - 6 } ^ { 6 } \int _ { - 8 } ^ { 8 } \sqrt { 1 + ( 2 x - 9 y ) ^ { 2 } + ( 18 y - 9 x ) ^ { 2 } } d x d y\)
• B)\(\int_{-8}^{8} \int_{-6}^{6} \sqrt{1+(2 x-9 y)^{2}+(18 y-9 x)^{2}} d y d x\)
• C) \(\int _ { - 6 } ^ { 6 } \int _ { - 8 } ^ { 8 } \sqrt { 1 + \left( 2 x ^ { 2 } - 9 y \right) ^ { 2 } + \left( 18 y ^ { 2 } - 9 x \right) ^ { 2 } } d y d x\)
• D)\(\int_{-6}^{6} \int_{-8}^{8} \sqrt{1+(2 x-9 y)^{2}+(18 y-9 x)^{2}} d y d x\)
• E)\(\int_{-6}^{6} \int_{-8}^{8} \sqrt{1+(2 x+9 y)^{2}+(18 y+9 x)^{2}} d y d x\)

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