Integrate the function $f$ over 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 $2 \leq x \leq 4$ and $f ( x , y ) = x ^ { 2 }$
A) $52 \sqrt { 83 }$
B) $26 \sqrt { 83 }$
C) $\frac { 52 } { 3 } \sqrt { 83 }$
D) $\frac { 26 } { 3 } \sqrt { 83 }$

Question

Quizplus · Accepted Answer

To integrate the function $f(x, y) = x^2$ over the surface $S$, which is a portion of the plane $z = x + 9y - 2$ above the rectangle $1 \leq x \leq 3$ and $2 \leq y \leq 4$, we first need to find the differential area element $dS$ on the surface. The surface can be parameterized as $\vec{r}(x, y) = \langle x, y, x + 9y - 2 \rangle$, and the normal vector $\vec{N}$ can be found by taking the cross product of the partial derivatives of $\vec{r}$ with respect to $x$ and $y$, which gives $\vec{N} = \langle 1, 9, -1 \rangle$. The magnitude of $\vec{N}$ is $\|\vec{N}\| = \sqrt{1^2 + 9^2 + (-1)^2} = \sqrt{83}$, so $dS = \sqrt{83} \, dx \, dy$.The integral of $f$ over $S$ is then given by$$\iint_S f \, dS = \iint_{R} x^2 \sqrt{83} \, dx \, dy$$where $R$ is the projection of $S$ onto the $xy$-plane, which is the rectangle $1 \leq x \leq 3$ and $2 \leq y \leq 4$.Evaluating the integral, we get$$\int_{1}^{3} \int_{2}^{4} x^2 \sqrt{83} \, dy \, dx = \sqrt{83} \int_{1}^{3} x^2 (4 - 2) \, dx = 2\sqrt{83} \int_{1}^{3} x^2 \, dx = 2\sqrt{83} \left[ \frac{x^3}{3} \right]_{1}^{3} = 2\sqrt{83} \left( \frac{27}{3} - \frac{1}{3} \right) = 2\sqrt{83} \cdot \frac{26}{3}.$$Thus, the correct answer is $\frac{52}{3} \sqrt{83}$, which corresponds to choice C.

Integrate the Function fff Over the Surface SSS Let SSS Be the Portion of the Plane

Integrate the Function $f$ Over the Surface $S$ Let $S$ Be the Portion of the Plane