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)$\int_{-6}^{6} \int_{-8}^{8} \sqrt{1+(2 x+9 y)^{2}+(18 y+9 x)^{2}} d y d x$

Question

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)$\int_{-6}^{6} \int_{-8}^{8} \sqrt{1+(2 x+9 y)^{2}+(18 y+9 x)^{2}} d y d x$

Quizplus · Accepted Answer

To find the surface area of the graph of $f$ over $R$, we can use the formula:
$$\int_R\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}dA$$
where $dA$ is the area element. Computing the partial derivatives, we have:
$$\frac{\partial f}{\partial x}=2x-9y \qquad \frac{\partial f}{\partial y}=-9x+18y$$
Plugging in these expressions, we get:
$$\int_{-8}^8\int_{-6}^6\sqrt{1+(2x-9y)^2+(18y-9x)^2}dxdy$$
which corresponds to option D.

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