Triple integration

**jcarlos** · August 22nd 2006, 01:42 PM

Hello,

Having problems with a homework question:

Part A asks to sketch the solid region bounded in the first octant by elliptic cylinder 2x^2 + y^2 =1, and the plane y + z = 1.

I know we have to do this question with cylindrical coordinates; thus, after substituting x = rcos(theta) and y = rsin(theta), I get:

z = 1- rsin(theta)

r = 2/(3 + cos2(theta)); after the substitution sin^2(theta) = 1 - cos^2(theta), and cos^2(theta) = (1 + cos2(theta))/2.

I have set up limits of integration as follows:

for d(theta), from 0 to pi/2 because of the first quadrant requirement
for dr from 0 to r = 2/(3 + cos2(theta))
for dz from 0 to z = 1- rsin(theta)

Of course I have not forgotten that dV = rdrd(theta)dz

I get stuck with an ugly trig expression when I get to the integration for d(theta)

What also confuses me is that the graph for the elliptic cylinder is not continuous, that is y = (1-2x^2)^(1/2) is not joined with
y = -(1-2x^2)^(1/2)!

Another quick quesiton:

How can you integrate cos^5(theta)d(theta) without looking it up in an integration table for which the answer also includes the integration of
cos^3(theta)d(theta)

For the integration of sin^3(theta)d(theta) I have seen the substituion of
u = cos(theta), thus, u^2 = cos^2(theta) = 1 - ( 1- sin^2(u)) =
sin^2(theta), and du = -sin(theta)d(theta). Therefore we can integrate
(1 - u^2)du which is equivalent to sin^3(theta)d(theta).

Is there a similar way for integrating cos^5(theta)d(theta)?

Thank you very much

**jcarlos** · August 22nd 2006, 02:40 PM

Moreover regarding my last question,

If I am to integrate (1- cos^5(theta))d(theta) from 0 to pi/2, can I split the integral into:
1d(theta) from 0 to pi/2 minus cos^5(theta)d(theta) from 0 to pi/2?
I then make the substitution u = sin(theta), thus, du = cos(theta)d(theta). Moreover, 1 - u^2 = 1 - sin^2(theta) = cos^2(theta). Therefore the integral for cos^5(theta)d(theta) from 0 to pi/2 is equivalent to integral of
(cos^2(theta))(cos^2(theta))cos(theta)d(theta) from 0 to pi/2 which is equivalent to integral of (1 - u^2)(1 - u^2)du from 0 to 1 (these new limits found by u = sin(theta) from 0 to pi/2).

By the described method above (splitting the integral) I get an answer of:
of (pi/2 - 8/15)

However when doing the same integral for 1- cos^5(theta))d(theta) from 0 to pi/2 using the table of integrals I get an anwer different from the above?!

Thanks again

**ThePerfectHacker** · August 22nd 2006, 02:45 PM

Originally Posted by jcarlos

Hello,

Having problems with a homework question:

Part A asks to sketch the solid region bounded in the first octant by elliptic cylinder 2x^2 + y^2 =1, and the plane y + z = 1.

Forgive me, I do not have a 3-D image grapher so I have to use MS-Paint.
Hope you understand.
(The first octant was shaded in red).

Originally Posted by jcarlos

Is there a similar way for integrating cos^5(theta)d(theta)?

Forgive me again. The answers might be out of order.

The answer is YES.
$\int \cos^5 xdx=\int (\cos^2 x)^2 \cos x dx$

$\int (1-\sin ^2x)^2 \cos x dx$
Use the functional substitution,
$u=\cos x$

**ThePerfectHacker** · August 22nd 2006, 03:14 PM

The volume of the region is,
$\int \int_S \int dV$
Using Fubini's theorem for space,
$\int_A \int \int_{v(x,y)}^{u(x,y)}dz\, dA$
Where,
$A$ is the area of the region presented below.
And,
$v(x,y)$ ---> lower curve, in this case zero.
$u(x,y)$ ---> upper curve, in this case $z=1-y$
Thus,
$\int_A \int \int_0^{1-y} dz\, dA=\int_A \int 1-y\, dA$

It is easier to use rectangular coordinate.
The region is a type I region.
The upper curve the equation,
$y=\sqrt{1-2x^2}$
The lower,
$y=0$
The x-bounds are,
$0 \leq x\leq \frac{\sqrt{2}}{2}$
Thus,
$\int_0^{\sqrt{2}/2} \int_0^{\sqrt{1-2x^2}} 1-y dy\, dx$
---
The smartest way to do this problem is to simplify the region by a change of variables and apply the Jacobian.

**ThePerfectHacker** · August 22nd 2006, 03:56 PM

Originally Posted by ThePerfectHacker

The smartest way to do this problem is to simplify the region by a change of variables and apply the Jacobian.

You have,
$2x^2+y^2=1$
Let,
$u=x\sqrt{2},v=y$ .
Then,
$2x^2+y^2=1\to u^2+v^2=1$ --->We turned an elliptical curve into a circular.
Then,
$x=u/\sqrt{2}$ and $y=v$
Then the Jacobian is,
$\frac{\partial(x,y)}{\partial(u,v)}=\left| \begin{array}{cc} \frac{\partial (u/\sqrt{2})}{\partial u}&\frac{\partial (u/\sqrt{2})}{\partial v}\\ \frac{\partial (v)}{\partial u}& \frac{\partial (v)}{\partial v} \end{array} \right|=\left| \begin{array}{cc}1/\sqrt{2}&0\\0&1 \end{array} \right|=1/ \sqrt{2}$
Therefore, (by change of variables)
$\int_{D^*} \int f(u,v) \left| \frac{\partial(x,y)}{\partial(u,v)} \right|\, dA$ = $\frac{1}{\sqrt{2}} \int_0^{2\pi} \int_0^1 r-r^2\sin \theta dr \, d\theta$

**galactus** · August 22nd 2006, 04:03 PM

I agree with PH, rectangular is best if possible. Here's the cylindrical if you must:

For the z limits with r and theta held fixed:

$z+y=1$

$z=1-rsin({\theta})$

For the r limits with z and theta held fixed:

$2x^{2}+y^{2}=1$

If you sub in $x=rcos({\theta})\;\ and\;\ y=rsin({\theta})$ and

solve for r you arrive at:

$r=\frac{1}{\sqrt{cos^{2}({\theta})+1}}$

Of course, the first octant is 0 to $\frac{\pi}{2}$

$\int_{0}^{\frac{\pi}{2}}$ $\int_{0}^{\frac{1}{\sqrt{cos^{2}({\theta})+1}}}$ $\int_{0}^{1-rsin({\theta})}rdzdrd{\theta}$

This will be hard to do by hand. I would use technology if available to evaluate.

**ThePerfectHacker** · August 22nd 2006, 04:15 PM

Originally Posted by galactus

I agree with PH, rectangular is best if possible. Here's the cylindrical if you must:

For the z limits with r and theta held fixed:

$z+y=1$

$z=1-rsin({\theta})$

For the r limits with z and theta held fixed:

$2x^{2}+y^{2}=1$

If you sub in $x=rcos({\theta})\;\ and\;\ y=rsin({\theta})$ and

solve for r you arrive at:

$r=\frac{1}{\sqrt{cos^{2}({\theta})+1}}$

Of course, the first octant is 0 to $\frac{\pi}{2}$

$\int_{0}^{2\pi}$ $\int_{0}^{\frac{1}{\sqrt{cos^{2}({\theta})+1}}}$ $\int_{0}^{1-rsin({\theta})}rdzdrd{\theta}$

This will be hard to do by hand. I would use technology if available to evaluate.

I encountoured (however you spell that) the same problem. At first, I assumed that I was making a error somewhere and that polar are ideal because it was a elliptic region. But you confirmed that it was not.

Of course, the ideal method is demonstrated in post #5.
When the elliptic region is changed to a circular.

**jcarlos** · August 22nd 2006, 09:42 PM

Gentlemen (PH and galactus)

Thank you very much.

Perfect Hacker, in reply 3, you mean let u = sinx and not u = cosx. If we let u =sinx, then du = cosxdx, that way we can integrate (1 - u^2)^2du.

To both of you,

I also did the question in rectangular as in reply 4. This makes sense as I got the same limits of integration. I get a final answer of (3pi -8)/(12(2)^1/2) = 0.0840 approx.

I don't understand changing the variables and applying Jacobian.

I have only done double integration for general regions as type I or II, polar coordinates, triple integration in general regions type I or II, circular coordinates. I have not got to spherical coordinates yet so anything past that point is above my head. However, you have made the Jacobian clear in reply 5 and I am able to understand it and see that it is a much easier integral to work with. I pressume that the rectangular and Jacobian methods should give the same volume right?!

Galactus in reply 7 you agree with PH that the best method is rectangular. I also agree with your cylindrical as that is what I have. You seem to integrate with respect to theta from 0 to 2pi instead of 0 to pi/2, is this a typo?
I have used quickmath.com to evaluate the iterated integral using cylindrical coordinates, I get to the last integral with respect to theta which I will name x for convenience right now and I get the integral expression of:
1/6[(3/cos^2(x) + 1) - sin(x)/(cos^2(x) + 1)^2 from x = 0 to x = pi/2. Remember I am using x in place for theta for convenience!
The computer program gives me a final answer in terms of many arctan functions which is simplified to an answer of 2.409, much higher than the answer I obtained from rectangular coordinates of 0.0840!

I will calculate the Jacobian method and get an answer but I need two clarifications first: Since it is the first octant we are integrating from 0 to pi/2 and not from 0 to 2pi, correct?! Also when integrating r - r^2sin(theta) with respect to r, sin(theta) is treated as a constant and thus we end up with r^2 - sin(theta)(r^3)/3 from r = 0 to r = 1.
I cannot help to notice that the Jacobian method simplifies the cylindrical method by doing a determinant and simplifying the expression for the upper limit of r to just r =1, right?

I really appreciate your help gentlemen. As you can see I am not slacking off but trying very hard. I am happy that your methods agree with what I am able to get.

**jcarlos** · August 22nd 2006, 10:30 PM

I did the Jacobian method integral and ended up with an answer of
(3pi -4)/(12(2)^(-1/2)) which does not agree with the rectangular answer
of (3pi -8)/(12(2)^(-1/2))!

I have doen both with quickmath software and by hand and I get the same answers which do not agree with each other!

I also integrated the Jacobian from 0 to 2pi and got a much different answer!

What is going on?

Thank you

**galactus** · August 23rd 2006, 03:18 AM

Yes, that was a typo. I fixed it. Thanks for pointing that out.

The answer for my cylindrical method is $\frac{(3{\pi}-4)\sqrt{2}}{24}\approx{0.319658}$

This agrees with PH's rectangular method.

**ThePerfectHacker** · August 23rd 2006, 06:36 AM

jcarlos do you understand the change of varibles (Jacobian) part? Should I re-explain.
---
A useful tool is whenever you are given an ellipse,
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
You can use the change of variables,
$\frac{x}{a}=u \mbox{ and }\frac{y}{b}=v$
That will make the elliptic region into a circle.
And the Jacobian itself is $ab$ a constant which make your work much simpler.
(Just remember to substitute the x for au and y for bv)

**jcarlos** · August 23rd 2006, 08:36 AM

Thank you again gentlemen,

Perfect Hacker, you were clear in expalining the change in variable for the Jacobian. The final integral you get is much simpler than the cylindrical method.
I did the Jacobian method integral and ended up with an answer of
(3pi -4)/(12(2)^(-1/2)) which does not agree with the rectangular answer
of (3pi -8)/(12(2)^(-1/2))!

I have done them both with quickmath software and by hand again and again. and I get the same answers which do not agree with each other!

Moreover, galactus gets an anwer for the cylindrical method of:
$<br /> \frac{(3{\pi}-4)\sqrt{2}}{24}\approx{0.319658}<br />$
Where he states that it agrees for perfect hackers rectangular method. I could not do the cylindrical method by hand and obviously the software program I used game me a wrong answer eventhough I checked the expressions and limits that I entered carefully. The bottom line is that galactus answer does not agree with my Jacobian and rectangular answers which also do not agree with each other!

This will be the last post for this subject; I am getting really frustrated. Thank you both again for your help!

**jcarlos** · August 23rd 2006, 11:20 AM

perfect hacker and galactus,

I have managed to get the same answer for rectangular as Jacobian.
Thank you both for your help.

I did not realize that 2^(-1/2) = (2^(1/2))/2.
Therefore, $<br /> \frac{(3{\pi}-4)\sqrt{2}}{24}\approx{0.319658}<br />$

Is equivalent to (3pi -4)/(12*(2^(1/2))

Thanks again!

**jcarlos** · August 23rd 2006, 11:23 AM

How do you spell 'differencial' = differential!

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