Evaluate the surface integral where $\sigma$ is that portion of the cylinder y² + z² = 1 that lies above the xy-plane enclosed by 0 $\le$ x $\le$ 5 and -1 $\le$ y $\le$ 1. E) 0

Question

Quizplus · Accepted Answer

First, we need to parameterize the surface. We can do this by using cylindrical coordinates:
$$x = r\cos	heta$$
$$y = y$$
$$z = r\sin	heta$$
where $0\leq r\leq1$ and $0\leq	heta\leq 2\pi$.

Next, we need to find the unit normal vector to the surface. Using the gradient:
$$
abla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} angle = \langle 0, 1, 0 angle$$
Thus, the unit normal vector is $\frac{1}{\sqrt{1+0+0}}\langle0,1,0angle = \langle0,1,0angle$.

To evaluate the surface integral, we use the formula:
$$\iint_S 	extbf{F}\cdot	extbf{n}\,dS = \int_0^{2\pi}\int_0^1 	extbf{F}(r\cos	heta, y, r\sin	heta)\cdot\langle0,1,0angle r\,dr\,d	heta$$
Plugging in $	extbf{F}$:
$$\iint_S y\,	extbf{n}\,dS = \int_0^{2\pi}\int_0^1 y\cdot1\cdot r\,dr\,d	heta$$
Integrating with respect to $r$:
$$\int_0^{2\pi}\int_0^1 y\cdot r\,dr\,d	heta = \int_0^{2\pi} \frac{1}{2}y\,d	heta = \pi y$$
Now we can evaluate the integral over the region given. By symmetry, we only need to consider the portion where $y\geq0$:
$$\int_0^1 \pi y\,dy = \frac{\pi}{2}$$
Therefore, the answer is C.

Evaluate the Surface Integral Where σ\sigmaσ Is That Portion of the Cylinder Y2 + Z2 =

Evaluate the Surface Integral Where $\sigma$ Is That Portion of the Cylinder Y² + Z² =