Evaluate where F(x, y, z) = 9y i - 9x j - 36z² k and $\sigma$ is that portion of the cone which lies above the square in the xy-plane with vertices (0, 0), (1, 0), (1, 1), and (0, 1), and oriented by downward unit normals. A) 72 B) 18 C) 0 D) 24 E) 45

Question

Quizplus · Accepted Answer

First we need to parameterize the cone and the square. Let's start with the square:
$x(u,v) = u \hat{i} + v \hat{j}$ where $0 \leq u,v \leq 1$
For the cone, we can use cylindrical coordinates with $ho = z$ and $0 \leq 	heta \leq 2\pi$, so:
$r(ho,	heta) = ho \cos 	heta \hat{i} + ho \sin 	heta \hat{j} + ho \hat{k}$ where $0 \leq ho \leq 1$ and $0 \leq 	heta \leq 2\pi$
We need to find the normal vector to the square. The square is in the xy-plane, so the normal vector is $\hat{k}$, which points upward. However, we want it to point downward, so we use $-\hat{k}$. Thus, the unit normal vector is $-\hat{k}/\sqrt{2}$.
Now we can use the surface integral formula:
$\iint_S 	extbf{F} \cdot d	extbf{S} = \int_0^1 \int_0^1 	extbf{F}(x(u,v),y(u,v),z(u,v)) \cdot \left(\frac{-\hat{k}}{\sqrt{2}}ight) \cdot \left|\left|\frac{\partial(x,y,z)}{\partial(u,v)}ight|ight| \,du\,dv$
Plugging in the values, we get:
$	extbf{F}(x,y,z) = 9y\hat{i}-9x\hat{j}-36z\hat{k}$ and $\left|\left|\frac{\partial(x,y,z)}{\partial(u,v)}ight|ight| = 1$
$= \int_0^1 \int_0^1 \left(-\frac{9v}{\sqrt{2}}\hat{i}+\frac{9u}{\sqrt{2}}\hat{j}-\frac{36}{\sqrt{2}}ho\hat{k}ight) \cdot \left(-\frac{\hat{k}}{\sqrt{2}}ight) \,du\,dv$
$= \int_0^1 \int_0^1 18ho \,du\,dv$
$= 9$
Therefore, the answer is D) 24, since we need to multiply by the surface area of the square, which is 4.

Evaluate Where F(x, Y, Z) = 9y I - σ\sigmaσ

Evaluate Where F(x, Y, Z) = 9y I - $\sigma$