surface integral - 2

logistic_guy

logistic_guy

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Evaluate the surface integral.

\(\displaystyle \iint\limits_S z \ dS\), where \(\displaystyle S\) is the portion of \(\displaystyle x^2 + y^2 = 1\) with \(\displaystyle x \geq 0\) and \(\displaystyle z\) between \(\displaystyle z = 1\) and \(\displaystyle z = 2\).
 
logistic_guy said:
Evaluate the surface integral.

\(\displaystyle \iint\limits_S z \ dS\), where \(\displaystyle S\) is the portion of \(\displaystyle x^2 + y^2 = 1\) with \(\displaystyle x \geq 0\) and \(\displaystyle z\) between \(\displaystyle z = 1\) and \(\displaystyle z = 2\).
Click to expand...
show us your effort/s to solve this problem.
 
Last edited:
khansaheb said:
show us your effort/s to solve this problem.
Click to expand...
👍

I have three methods to solve this problem.

First Method:

My strategy to solve this problem is first to understand or to visualize the portion of the surface area \(\displaystyle S\). Well it is just a half cylinder with height \(\displaystyle 1\) rises in the \(\displaystyle z\) direction.

From simple geometry, I know that the surface area of a cylinder is \(\displaystyle 2\pi r h + 2\pi r^2\),

where \(\displaystyle 2\pi r h\) is the lateral surface area and \(\displaystyle 2\pi r^2\) is the circular surface area of the top and the bottom of the cylinder.

This problem is only interested in the lateral surface area, i.e., the curved surface around the sides. If the full curved surface area is \(\displaystyle S = \pi rh\), a very small portion of this area will be \(\displaystyle dS = \pi r \ dz\). (Remember that we are working only with a half cylinder.)

I have taken the differential element \(\displaystyle dz\) in this case because the height of the cylinder is rising in the \(\displaystyle z\) direction.

Then

\(\displaystyle \iint\limits_S z \ dS = \int_{1}^{2}z \pi r \ dz = \pi (1)\int_{1}^{2} z \ dz = \pi \frac{z^2}{2}\bigg|_{1}^{2} = \pi \left(\frac{2^2}{2} - \frac{1^2}{2}\right)\)

\(\displaystyle = \pi \left(\frac{4}{2} - \frac{1}{2}\right) = \pi \left(\frac{3}{2}\right) = \frac{3\pi}{2}\)

That was not the interesting way to solve the problem. You will have to wait until you see Method 3.😉
 
Method 2

Differential geometry approach. This method depends on parametrizing the surface. While it is not the fun method to solve the problem, it is the core method in solving all types of surface integrals.

Let us first write the surface integral in a different form.

\(\displaystyle \iint\limits_S z \ dS = \int\int z \ |\bold{n}| \ d\theta \ dz\)

where \(\displaystyle \bold{n}\) is the normal vector.

We have a cylinder, so the best parametrization is to use the cylindrical coordinate.

\(\displaystyle x = r\cos \theta\)
\(\displaystyle y = r\sin \theta\)
\(\displaystyle z = z\)

Since the radius \(\displaystyle r = 1\), this will be reduced to:

\(\displaystyle x = \cos \theta\)
\(\displaystyle y = \sin \theta\)
\(\displaystyle z = z\)

Then, the parametrization of the surface \(\displaystyle S\) is:

\(\displaystyle \boldsymbol{ \varphi}(\theta,z) = (\cos \theta, \sin \theta, z)\)

The normal vector is just the cross product of the partial derivatives of our parametrized surface.

\(\displaystyle \bold{n} = \frac{\partial \boldsymbol{\varphi}}{\partial \theta} \times \frac{\partial \boldsymbol{\varphi}}{\partial z}\)

With a little calculation, you will find that \(\displaystyle \frac{\partial \boldsymbol{\varphi}}{\partial \theta} \times \frac{\partial \boldsymbol{\varphi}}{\partial z} = 1\)

Then

\(\displaystyle \int\int z \ |\bold{n}| \ d\theta \ dz = \int_{1}^{2}\int_{0}^{\pi} z \ |1| \ d\theta \ dz\)

\(\displaystyle = \int_{1}^{2}\pi z \ dz = \frac{3\pi}{2}\)

One more step to see Method 3😍
 
Method 3

Finally😻

Simple calculus approach. While this method inherited its power from method 2, it is the most fun method to solve surface integrals. Besides it's an entertaining method, it has the advantage of that we don't have to deal with vectors and cross products headaches.

This method only requires you to sharpen your pencil and show the professor that you are the boss of differentiation and integration techniques.

Let us first write the integral in a different form.

\(\displaystyle \iint\limits_S z \ dS = \int\int z \sqrt{(f_y)^2 + (f_z)^2 + 1} \ dy \ dz\)

where \(\displaystyle f_y\) and \(\displaystyle f_z\) are just the partial derivatives of \(\displaystyle x = f(y,z) = \sqrt{1 - y^2}\)

\(\displaystyle \int\int z \sqrt{(f_y)^2 + (f_z)^2 + 1} \ dy \ dz = \int_{1}^{2}\int_{-1}^{1} z \sqrt{\frac{y^2}{1 - y^2} + 0 + 1} \ dy \ dz\)

\(\displaystyle = \int_{1}^{2} z \pi \ dz = \pi\frac{z^2}{2}\bigg|_{1}^{2} = \frac{3\pi}{2}\)

Solving the first integral looks complicated, but if you look closer it is just a simple \(\displaystyle u\) substitution. I'll explain how to solve

\(\displaystyle \int_{-1}^{1}\sqrt{\frac{y^2}{1 - y^2} + 1} \ dy\)

in the next post. This method was fun, wasn't it?😉I told you!
 
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Let us crack that integral with some steps.

\(\displaystyle \int_{-1}^{1}\sqrt{\frac{y^2}{1 - y^2} + 1} \ dy\)

First step simplify.

\(\displaystyle = 2\int_{0}^{1}\sqrt{\frac{y^2}{1 - y^2} + \frac{1 - y^2}{1 - y^2}} \ dy = 2\int_{0}^{1}\frac{1}{\sqrt{1 - y^2}} \ dy\)

Let
\(\displaystyle y = \sin u\)
\(\displaystyle dy = \cos u \ du\)

\(\displaystyle = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos u}{\sqrt{1 - \sin^2 u}} \ du = 2\int_{0}^{\frac{\pi}{2}} du = 2\frac{\pi}{2} = \pi\)

😎
 
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