# Surface integral over a cylinder bounded by 2 planes

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.

• Try using polar coordinates i.e. $x=r cos\theta$, $z=r sin\theta$ and integrate over $\theta$ and $r$.
– user473029
Dec 27 '18 at 15:15
• I tried, failed again. Dec 27 '18 at 15:20
• It's hard to answner this without the Divergence Theorem [...] Try to use the Stoke's theorem (Surface integral into a integral over the boundary).
– user473029
Dec 27 '18 at 16:56
• Thank you for the alternative, I found an answer at math.stackexchange.com/questions/2222773/… Dec 27 '18 at 17:46
• Your calculations are correct, the integral that you've obtained is $\iint_R 5 x \,dx dy = \int_0^3 \int_0^8 5 x \,dy dx$. Now you need to compute the integrals over the other parts of the surface. Dec 27 '18 at 17:52

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

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

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