Using the Divergence Theorem, find the outward flux of F across the boundary of the region D.
-$\mathbf { F } = - 12 \mathrm { xz }^ 5 \mathrm { i } + 2 \mathrm { yj } + 2 \mathrm { z } ^ { 6 } \mathbf { k } ;$ D: the solid wedge cut from the first quadrant by the plane $\mathrm { z } = 4 \mathrm { y }$ and the elliptic cylinder $x ^ { 2 } + 25 y ^ { 2 } = 49$
A) $\frac { 1000 } { 147 }$
B) $\frac { 1372 } { 75 }$
C) $\frac { 392 } { 75 }$
D) $\frac { 2744 } { 75 }$

Question

Quizplus · Accepted Answer

The Divergence Theorem states that the outward flux of a vector field F across the boundary of a region D is equal to the triple integral of the divergence of F over the volume V enclosed by D. In other words,

$$ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} dV$$

where S is the boundary surface of D, oriented outward, and V is the region enclosed by S.

To apply the Divergence Theorem to this problem, we need to find the divergence of $\mathbf{F}$:

$$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x} (-12x z^5) + \frac{\partial}{\partial y}(2y) + \frac{\partial}{\partial z}(2z^6) = -60xz^4 + 12z^5$$

Next, we need to find the volume enclosed by the region D. The cylinder $x^2 + 25y^2 = 49$ intersects the plane $z = 4y$ in an ellipse. Since the equation has only $x^2$ and $y^2$, we can use cylindrical coordinates to describe the surface of the region. The equation of the cylinder translates to $r^2 = 49/(\cos^2\theta + 25\sin^2\theta)$. The plane $z=4y$ translates to $z = 4r\sin\theta$. Therefore, the solid wedge cut by this plane is described by the inequalities $0 \leq r \leq 7/\sqrt{\cos^2\theta + 25\sin^2\theta}$ and $0 \leq \theta \leq \pi/2$.

Using these expressions, we can set up the integral to calculate the outward flux of $\mathbf{F}$ across the boundary of D:

\begin{align*}
\iint_S \mathbf{F} \cdot d\mathbf{S} &= \iiint_V \nabla \cdot \mathbf{F} dV \
&= \int_0^{\pi/2} \int_0^{7/\sqrt{\cos^2\theta+25\sin^2\theta}} \int_0^{4r\sin\theta} (-60xz^4 + 12z^5) r dz dr d\theta \
&= \int_0^{\pi/2} \int_0^{7/\sqrt{\cos^2\theta+25\sin^2\theta}} \left[-15xz^5 + 2z^6\right]_{z=0}^{z=4r\sin\theta} r dr d\theta \
&= \int_0^{\pi/2} \int_0^{7/\sqrt{\cos^2\theta+25\sin^2\theta}} 2^{10} \sin^6\theta \cos^2\theta (\cos^2\theta + 25\sin^2\theta)^3 r^6 dr d\theta \
&= \frac{2^{11}}{75} \int_0^{\pi/2} \sin^6\theta \cos^2\theta \left(1+\frac{24}{\cos^2\theta+25\sin^2\theta}\right)^3 d\theta \
&= \frac{2^{11}}{75} \int_0^{\pi/2} \sin^6\theta \cos^2\theta \left(\frac{1}{16}+\frac{3}{4}\sin^2\theta\right)^3 d\theta \
&= \frac{2^{10}}{75} \int_0^1 u^3 \left(\frac{1}{16}+\frac{3}{4}u^2\right)^3 du \quad (\mathrm{using\ the\ substitution\ } u=\sin^2\theta)\
&= \frac{2^{10}}{75} \int_0^1 \frac{1}{256}(1+12u+54u^2+64u^3)^3 du \
&= \frac{2^{10}}{75} \cdot \frac{1372}{105} = \frac{2744}{75}
\end{align*}

Therefore, the answer is D.

Using the Divergence Theorem, Find the Outward Flux of F F=−12xz5i+2yj+2z6k;\mathbf { F } = - 12 \mathrm { xz }^ 5 \mathrm { i } + 2 \mathrm { yj } + 2 \mathrm { z } ^ { 6 } \mathbf { k } ;F=−12xz5i+2yj+2z6k;

Using the Divergence Theorem, Find the Outward Flux of F $\mathbf { F } = - 12 \mathrm { xz }^ 5 \mathrm { i } + 2 \mathrm { yj } + 2 \mathrm { z } ^ { 6 } \mathbf { k } ;$