$\text { Find the area of the portion of the surface } z = 8 x + 4 y \text { that lies above the triangular }$ region with vertices $( 0,0 ) , ( 2,0 )$, and $( 0,2 )$.
A) 4
B) 36
C) 18
D) 162
E) 324

Question

$\text { Find the area of the portion of the surface } z = 8 x + 4 y \text { that lies above the triangular }$  region with vertices $( 0,0 ) , ( 2,0 )$, and $( 0,2 )$.
A) 4
B) 36
C) 18
D) 162
E) 324

Quizplus · Accepted Answer

The area of a surface $z = f(x, y)$ above a region in the $xy$-plane can be found using the formula $\iint_{D} \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \,dA$, where $D$ is the projection of the surface onto the $xy$-plane. For $z = 8x + 4y$, we have $\frac{\partial z}{\partial x} = 8$ and $\frac{\partial z}{\partial y} = 4$. Substituting these into the formula gives $\iint_{D} \sqrt{1 + 8^2 + 4^2} \,dA = \iint_{D} \sqrt{81} \,dA = 9\iint_{D} dA$. The region $D$ is a right triangle with legs of length 2, so its area is $\frac{1}{2} \cdot 2 \cdot 2 = 2$. Therefore, the surface area is $9 \cdot 2 = 18$.

Find the area of the portion of the surface z=8x+4y that lies above the triangular \text { Find the area of the portion of the surface } z = 8 x + 4 y \text { that lies above the triangular } Find the area of the portion of the surface z=8x+4y that lies above the triangular

$\text { Find the area of the portion of the surface } z = 8 x + 4 y \text { that lies above the triangular }$