Volume and Integration

By Hal 2001 (P3046) on Friday, October 13, 2000 - 01:30 pm :

Hi there,

I have tried without luck to solve the following problem.

The Question:

The finite region bounded by the curve

y = \tan (x / 2)

, the line

x = (π / 2)

and the coordinate axis is rotated through

2 π

radians about the

x

-axis. Show that the volume generated is

(π / 2) \times (4 - π)

.
What I have done so far...
We know that: Vol

= π \int y^{2} dx

So the vol for the question is: Vol $= π \int \tan^{2} (x / 2) dx$
But how do you go about integrating the previous line?

My ideas...
Do you use $1 + \tan^{2} x = \sec^{2} x$ ?

And then change it to: $1 + \tan^{2} (x / 2) = \sec^{2} (x / 2)$ ?

So that you get: Vol $= π \int \sec^{2} (x / 2) - 1 dx$ ,

Then once you integrate the above you get Vol $= π (2 \tan (x / 2) - x)$ ??
Help warmly appreciated.
Signing Out...
HAL2001

PS: Be kind, this is my 1st math related post!

By Richard Samworth (Rjs57) on Friday, October 13, 2000 - 06:26 pm :

Dear Hal,

You're so nearly there! You did the integral correctly, and now just need to evaluate the result at the upper limit of integration (i.e. $x = π / 2$ ) and subtract the expression evaluated at the lower limit of integration (i.e. $x = 0$ ). This gives you the answer you were looking for.

If you wanted to do the integral another way (or wanted to check your answer), you could use the substitution $y = \tan (x / 2)$ - try it and see!

Please write back if you don't understand why I've done what I have.

Best wishes

Richard

By Hal 2001 (P3046) on Friday, October 13, 2000 - 08:28 pm :

Hi Richard,

Thank you for your positive response to my problem. It is very much appreciated!

Further progress

Firstly, Vol

= π [2 \tan (x / 2) - x]_{0}^{π / 2}

Then: Vol $= π [((2 \tan (π / 4) - (π / 2) 0 - (2 \tan 0 - 0)]$ as required.

Hence: $π [2 - (π / 2)] = 2 π - (π^{2} / 2) = (π / 2) \times (4 - π)$ as required.
Is this OK?

Richard, could I have approached the integration
differently instead of using the identity I used?
Does there exist an easier approach to this problem?

Thanks again.
HAL2001

By Richard Samworth (Rjs57) on Saturday, October 14, 2000 - 05:20 pm :

Hal,

Well done - that's all correct! When I tried the problem from scratch I used the substitution y = tan(x/2) to evaluate the integral, which gave the same answer, but was perhaps not quite as elegant as your method. Please ask if you are not familiar with integration by substitution.

In general with these 'volumes of revolution' questions, except in simple cases, you are going to have to do the integral given in your formula and my feeling about your particular question is that while other substitutions into the integral might well give the answer, they are unlikely to make it any easier than your method.

One simple case where you wouldn't need to bother doing an integral would be if the finite region were that between the positive x-axis, the line y=x and the line x=1, and the question were to find the volume enclosed when this region were rotated through 2pi radians around the x-axis. In this case, the volume is just that of a cone with base radius 1 unit and 'vertical' height 1 unit. Of course, if you couldn't remember the formula for the volume of a cone, you could always do the integral using your formula just as in your question!

I hope this helps.

Richard

By Hal 2001 (P3046) on Monday, October 16, 2000 - 12:53 pm :

Hi Richard,

Thanks for your advice. It is much appreciated. :)

I tried to do it with your sub y = tan(x/2), but I did'nt get the same result! Is it possible for you to put your method workings up?

Regarding the volume of a cone, to get it into the form

V = (π r^{2} h / 3)

, what boundary conditions do you use?

x = 0

and

x = h

?
Regards
HAL2001

By Richard Samworth (Rjs57) on Tuesday, October 17, 2000 - 10:59 pm :

Hal,

When making a substitution into an integral, there are three things to do:

1) Write the integrand (the expression you're integrating) in terms of the new variable

2) Find how an infinitessimal change in the new variable is related to an infinitessimal change in the old one (i.e. ''find $dx$ in terms of $dy$ '')

3) Alter the limits of integration so that they apply to the new variable.

So here we go: we want to evaluate $π \int_{0}^{π / 2} \tan^{2} (x / 2) dx$ by means of the substitution $y = \tan (x / 2)$ .

In step 1, the integrand becomes $y^{2}$ . We do step 2 by noting that

$dy = 1 / 2 \times \sec^{2} (x / 2) dx = 1 / 2 \times (1 + \tan^{2} (x / 2)) dx = 1 / 2 \times (1 + y^{2}) dx$ .

Finally, we do step 3. When $x = 0$ , $y = 0$ , and when $x = π / 2$ , $y = 1$ .

Putting all these steps together, we end up with:

$Vol = 2 π \int_{0}^{1} y^{2} / (1 + y^{2}) dy = 2 π \int_{0}^{1} 1 - 1 / (1 + y^{2}) dy$ .

The second term in this integral is a standard integral - it becomes $\arctan y$ (the inverse of the tan function).

Hence the volume is

$Vol = 2 π [y - \arctan y]_{0}^{1} = 2 π (1 - π / 4) = π / 2 \times (4 - π)$ , as required.

I hope this makes some sense, and that you can see where your working differed. Finally, you're right about the boundary conditions for the cone. You need $x = 0$ and $x = h$ because this gives you the 'vertical' height of $h$ .

Again, please write back if there's anything you don't understand.

Best wishes

Rich

By Hal 2001 (P3046) on Wednesday, October 18, 2000 - 03:12 pm :

Thanks Richard,

Your method was very enlightening.
I enjoyed it very much!

The step where you integrated the arctan y, I did not remember from memory, so I checked up in a text book.

About the volume of the cone, I put

x = 0

and

x = h

as the boundary conditions, but got

(π \times h^{3}) / 3

instead of the correct answer we expect of

V = (π \times r^{2} h) / 3

.
Any suggestions on how I went wrong?

Best Regards
HAL2001

By Kerwin Hui (Kwkh2) on Wednesday, October 18, 2000 - 04:52 pm :

HAL,

I suspect you have used the equation $y = x$ instead of the correct equation $y = r x / h$ in your calculation. It is quite easy to see that you want to have $y = r$ when $x = h$ and $y = 0$ when $x = 0$ , where $r$ is the radius of the base. In using $y = x$ , we are having a cone with its radius equal to its height and thus yields $π h^{3} / 3$ .

If my suspicion is wrong, then there must be some mistakes in the evaluation of integral...

Kerwin

By Hal 2001 (P3046) on Wednesday, October 18, 2000 - 10:54 pm :

Hi Kerwin,

Thanks!
I used y=rx/h and it worked out.

Regards
HAL2001