Surface integral over a cylinder bounded by 2 planes

Question

From Schaum's vector analysis:

My attempt:

$\vec n = \nabla S = 2x \hat i + 2z \hat k$

$ \hat n = \frac{1}{3} x \hat i + \frac{1}{3} z \hat k $

$ \vec A . \hat n = 2xz - \frac{xz}{3} = \frac {5}{3} xz$

$dS = \frac {dxdy}{ \hat n . \hat k}$ , $ \hat n . \hat k = \frac {z}{3}$

$dS = \frac {3}{z} dxdy$

$ \iint_S \vec A . \hat n dS = 5 \iint_R x dxdy$

I know $y$ ranges from $0$ to $8$ then

$5 \iint_R x dxdy = 40 \int x dx$

This is where I stop, I can't integrate $x$ from $0$ to $3$ directly and I can't substitute it with the equation $x^2 + z^2 = 9$ , how do I continue? Also without making use of the divergence theorem please.

Answer 1

The surface $S$ has 5 distinct parts, so

$$ \int_S {\bf A}\cdot {\rm d}^2{\bf S} = \sum_{k=1}^5\int_{S_k} {\bf A}\cdot {\rm d}^2{\bf S} $$

where

$S_1 = \{(x, y, z)| x = 3\cos \theta, z = 3\sin\theta, 0\leq y \leq 8,0\leq \theta \leq \pi/2 \}$ (part of the cylinder)

The surface differential for this case is ${\rm d}^2{\bf S} = (3\cos\theta \hat{x} + 3\cos\theta \hat{z}){\rm d}\theta {\rm d}y$, so the integral becomes

$$ \int_{S_1} {\bf A}\cdot {\rm d}^2{\bf S} = 45\int_0^{\pi/2}{\rm d}\theta\int_0^8{\rm d}y \sin\theta\cos\theta = 180 \tag{1} $$

$S_2 = \{(x, y, z) | y = 0, x = r\cos\theta, z = r\sin\theta, 0\leq r \leq 3, 0\leq \theta \leq \pi/2 \}$ (face at $y = 0$)

In this case ${\rm d}^2{\bf S} = -\hat{y}r{\rm d}r{\rm d}\theta$ and

$$ \int_{S_2} {\bf A}\cdot {\rm d}^2{\bf S} = -2\int_0^3{\rm d}r\int_0^{\pi/2}{\rm d}\theta r^2\cos\theta = -18 \tag{2} $$

$S_3 = \{(x, y, z) | y = 8, x = r\cos\theta, z = r\sin\theta, 0\leq r \leq 3, 0\leq \theta \leq \pi/2 \}$ (face at $y = 8$)

In this case ${\rm d}^2{\bf S} = +\hat{y}r{\rm d}r{\rm d}\theta$ and

$$ \int_{S_3} {\bf A}\cdot {\rm d}^2{\bf S} = \int_0^3{\rm d}r\int_0^{\pi/2}{\rm d}\theta r(8 + 2r\cos\theta) = 18(1 + \pi) \tag{3} $$

$S_4 = \{(x, y, z) | z = 0, 0\leq x \leq 3, 0\leq y \leq 8 \}$ (face at $z = 0$)

In this case ${\rm d}^2{\bf S} = -\hat{z}{\rm d}x{\rm d}y$ and

$$ \int_{S_4} {\bf A}\cdot {\rm d}^2{\bf S} = \int_0^3{\rm d}x\int_0^{8}{\rm d}y x = 36 \tag{4} $$

$S_5 = \{(x, y, z) | x = 0, 0\leq z \leq 3, 0\leq y \leq 8 \}$ (face at $x = 0$)

In this case ${\rm d}^2{\bf S} = -\hat{x}{\rm d}z{\rm d}y$ and

$$ \int_{S_5} {\bf A}\cdot {\rm d}^2{\bf S} = -\int_0^3{\rm d}z\int_0^{8}{\rm d}y 6x = -216 \tag{5} $$

Adding all up

The result is

$$ \int_S {\bf A}\cdot {\rm d}^2{\bf S} = \sum_{k=1}^5\int_{S_k} {\bf A}\cdot {\rm d}^2{\bf S} = 180 - 18 + 18(1 + \pi) + 36 - 216 = \color{red}{18\pi} $$

Answer 2

This is a Divergence Theorem problem. The surface integral is equal to the the triple integral over the solid of the divergence of the vector field. Since the divergence equals $1$, the answer is the volume of the quarter cylinder (which is $18\pi.$