Evaluate the surface integral. S is the part of the plane that lies in the first octant.

Question

Evaluate the surface integral.  S is the part of the plane  that lies in the first octant.

Quizplus · Accepted Answer

The given surface integral can be written as:
$$\iint_S (2x+4y+6z) \ dS$$
where $S$ is the part of the plane $2x+4y+6z=12$ that lies in the first octant. Solving for $z$ in the equation of the plane, we get:
$$z=\frac{1}{2}-\frac{1}{3}x-\frac{1}{6}y$$
Since $S$ lies in the first octant, we have $0 \leq x \leq 6$, $0 \leq y \leq 2(6-x)$ and $0 \leq z \leq \frac{1}{2}-\frac{1}{3}x-\frac{1}{6}y$. Therefore, the surface integral becomes:
\begin{align*}
\iint_S (2x+4y+6z) \ dS &= \int_0^6 \int_0^{2(6-x)} (2x+4y+6z)\sqrt{\bigg(\frac{\partial z}{\partial x}\bigg)^2+\bigg(\frac{\partial z}{\partial y}\bigg)^2+1} \ dy \ dx \
&= \int_0^6 \int_0^{2(6-x)} (2x+4y+6\bigg(\frac{1}{2}-\frac{1}{3}x-\frac{1}{6}y\bigg))\sqrt{\bigg(-\frac{1}{3}\bigg)^2+\bigg(-\frac{1}{6}\bigg)^2+1} \ dy \ dx \
&= \int_0^6 \int_0^{2(6-x)} (5x+3y-3) \ dy \ dx \
&= \int_0^6 \bigg(\frac{3}{2}y+\frac{5}{2}xy-\frac{3}{2}y^2\bigg)\biggvert_{y=0}^{y=2(6-x)} \ dx \
&= \int_0^6 (27x-18x^2) \ dx \
&= 54\int_0^6 (3x-x^2) \ dx \
&= 54\bigg[\frac{3}{2}x^2-\frac{1}{3}x^3\bigg]_0^6 \
&= 432
\end{align*}
Therefore, the correct answer is (B).

Evaluate the Surface Integral