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.
A) $I = 37$
B) $I = 47$
C) $I = 32$
D) $I = 57$
E) $I = 67$

Question

Quizplus · Accepted Answer

We first need to find the limits of integration. Since the plane lies in the first octant, we have: $$0 \leq x \leq 2$$ $$0 \leq y \leq 2-x$$ $$0 \leq z \leq 4-2x-2y$$ Next, we need to find the unit normal vector to the surface S. This can be done by taking the partial derivatives of the equation of the plane: $$\vec{n} = \left<2, 2, -1\right>$$ We can normalize this vector to get the unit normal vector: $$\hat{n} = \frac{\vec{n}}{\left\|\vec{n}\right\|} = \frac{1}{3}\left<2, 2, -1\right>$$ Finally, we can evaluate the surface integral: $$\iint _ { S } 8 x z d S = \int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 2-x } \int _ { 0 } ^ { 4-2x-2y } 8xz \frac{1}{3}\left<2, 2, -1\right> \cdot d\vec{S}$$ We can simplify this by taking the dot product: $$\frac{8}{3} \int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 2-x } \int _ { 0 } ^ { 4-2x-2y } xz(2+2-y) \,dz\,dy\,dx$$ Evaluating the triple integral gives us: $$\frac{8}{3} \int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 2-x } \frac{(2-x-y)(4-2x-2y)^2}{2} \,dy\,dx = \frac{8}{3} \int _ { 0 } ^ { 2 } \int _ { 0 } ^ { 2-x } (-2x^2-6xy+10x+7y^2-14y+8)\,dy\,dx = \frac{32}{3}$$ Therefore, the answer is C.

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

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