Find the area of the surface $S$ Let $S$ be the portion of the plane $z = x + 9 y - 2$ above the rectangle $1 \leq x \leq 3$ and $2 \leq y \leq 4$
A) $4 \sqrt { 83 }$
B) $2 \sqrt { 83 }$
C) $\sqrt { 83 }$
D) $\frac { 1 } { 2 } \sqrt { 83 }$

Question

Quizplus · Accepted Answer

The area of the surface $S$ can be found using the formula for the area of a surface defined by $z = f(x, y)$ over a region $R$ in the $xy$-plane: $A = \int \int_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA$. For $z = x + 9y - 2$, $\frac{\partial z}{\partial x} = 1$ and $\frac{\partial z}{\partial y} = 9$, so the integrand becomes $\sqrt{1 + 1^2 + 9^2} = \sqrt{83}$. The area of the rectangle is $(3-1)(4-2) = 4$, so the total area is $4\sqrt{83}$.

Find the Area of the Surface SSS Let SSS Be the Portion of the Plane

Find the Area of the Surface $S$ Let $S$ Be the Portion of the Plane