Evaluate the triple integral

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April 16th 2008, 12:36 PM
Somersault

Evaluate the triple integral

Hi,

So my task is to evaluate this triple integral:

SSS yz DV, over the region E: above z = 0, below z = y, and inside the cylinder x^2 + y^2 = 4.

I thought it'd be best to use cylindrical coordinates. Thus:

0 <= z <= rsinθ
0 <= r <= 2
0 <= θ <= 2π

But I keep getting 0. The answer is actually 64/15.

Here are the steps I took to compute the integral:

= SSS r^2 * sinθ * z dz dr dθ
= SS r^2 * sinθ * [(z^2)/2] from 0 to rsinθ .. dr dθ
= SS (r^4 * (sinθ)^3)/2 dr dθ
= S ((sinθ)^3)/2 * [(r^5)/5] from 0 to 2 dθ
= (16/5) S ((sinθ)^3) dθ
use the trig identity to split it up (sinθ)^2 = 1 - (cosθ)^2, the use substitution... etc

the final result:

(16/5) * [((cosθ)^3)/3 - cosθ] from 0 to 2π

This is obviously zero, since cos(0) and cos(2π) produce the same result. What am I doing wrong?
April 16th 2008, 12:50 PM
Moo

Hello,

Did you multiply by the determinant of the Jacobian matrix when changing to cylindrical coordinates ?
April 16th 2008, 12:52 PM
Somersault

Quote:

Originally Posted by Moo

Hello,

Did you multiply by the determinant of the Jacobian matrix when changing to cylindrical coordinates ?

No, we have not covered that material. This problem should be able to be solved without using that method.
April 16th 2008, 12:58 PM
Moo

Ok, sorry...

How do you get [r² sin(theta) z] from yz then ? :)
April 16th 2008, 01:00 PM
Somersault

Quote:

Originally Posted by Moo

Ok, sorry...

How do you get [r² sin(theta) z] from yz then ? :)

y = r sinθ
z = z

Oh, whoops.. I guess I did multiply it by the determinant of the Jacobian matrix... I really should start looking up the proper definitions of things... I didn't exactly know the name of it... and of course, this is just a plug-and-chug calculus course... so the lecturers dont bother to explain anything rather than just say "you must do $BLAH to do this"

So...

rsinθ * z * r * dz dr dθ
April 16th 2008, 01:10 PM
galactus

Let's do it in rectangular:

$\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}(4-x^{2}-y^{2})dzdydx=\frac{64}{15}$

Now, convert this to cylindrical or even spherical.
April 16th 2008, 01:12 PM
Moo

But I still don't understand why there is a mistake... I've done twice the calculus without finding the problem :(

Sorry for the "Jacobian", I don't know how you call it..
April 16th 2008, 01:15 PM
Somersault

Quote:

Originally Posted by galactus

Let's do it in rectangular:

$\int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}(4-x^{2}-y^{2})dzdydx=\frac{64}{15}$

Now, convert this to cylindrical or even spherical.

Could you explain the steps involved? I don't understand how this is equivalent to the original integral:

$\int_{}\int_{}\int_{} yz dz$
April 16th 2008, 01:26 PM
galactus

Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
In that event:

$\int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}$

Same thing.
April 16th 2008, 01:28 PM
Moo

Quote:

Originally Posted by galactus

Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
In that event:

$\int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}$

Same thing.

I find this equal to 0... (Thinking)
April 16th 2008, 01:34 PM
Somersault

Quote:

Originally Posted by galactus

Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
In that event:

$\int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}$

Same thing.

Why this:

$0 <= y <= \sqrt{4-x^{2}}$ instead of $-\sqrt{4-x^{2}} <= y <= \sqrt{4-x^{2}}$

Because z has to be above 0?
April 16th 2008, 01:40 PM
Somersault

Solving it with Cartesian coordinates works... but I don't understand why doing it in polar does not... is it because theta should be between 0 and pi and not 0 and 2pi?

EDIT: Yep! That's. D'oh! What a stupid mistake... It's because z is bounded by 0 and y... and if z's lower bound is 0, you obviously can only have the top half of the circle in the xy-plane... =) which is from 0 to pi
April 16th 2008, 01:47 PM
galactus

Here's the polar set up:

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{rsin {\theta}}r^{3}sin^{2}{\theta} \;\ dzdrd{\theta}$

Next, try spherical.
April 16th 2008, 01:49 PM
galactus

Quote:

Originally Posted by Moo

I find this equal to 0... (Thinking)

No, it's 64/15.
April 16th 2008, 01:50 PM
galactus

Quote:

Originally Posted by Somersault

Why this:

$0 <= y <= \sqrt{4-x^{2}}$ instead of $-\sqrt{4-x^{2}} <= y <= \sqrt{4-x^{2}}$

Because z has to be above 0?

Then use the negative portion, you'll get the same result.

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