Evaluate the surface integral.
$$\iint _ { S } 8 x z d S$$
$S$ is the part of the plane $2 x + 2 y + z = 4$ that lies in the first octant. Select the correct answer.
A) $I = 67$
B) $I = 37$
C) $I = 32$
D) $I = 47$
E) $I = 57$

Question

Quizplus · Accepted Answer

We can parameterize the surface using $x$ and $y$ as:
$x=x$
$y=y$
$z=4-2x-2y$
The normal vector to this surface is $\vec{n}=\langle 2, 2, -1angle$. Using the formula for surface integrals, we get:
$$\iint_S 8xz dS=\iint_R8x(4-2x-2y)dA$$
where $R$ is the projection of $S$ onto the $xy$-plane. This is equivalent to finding the double integral over the region $R$:
$$\iint_R (32x-16x^2-16xy)dA$$
We can then solve this using iterated integrals:
$$\int_0^2\int_0^{2-x}(32x-16x^2-16xy)dydx=\int_0^2\left[32xy-16x^2y-8y^2xight]_{y=0}^{y=2-x}dx$$
$$=\int_0^2(32x-48x^2+16x^3-8(2-x)^2x)dx=\int_0^2(-8x^3-32x^2+32x)dx$$
$$=[-2x^4-32/3x^3+16x^2]_0^2=32/3\approx 10.6667$$
Therefore, the answer is closest to $\boxed{	extbf{(C) }\,I = 32}$.

Evaluate the Surface Integral ∬S8xzdS\iint _ { S } 8 x z d S∬S​8xzdS

Evaluate the Surface Integral $\iint _ { S } 8 x z d S$