What is the Correct Result for This Triple Integral Problem?

[Bassalisk]

Homework Statement

I=(triple intgral)(x²+y²)dxdydz.

D:
z=2
x²+y²=2z
z>=0

Homework Equations

The Attempt at a Solution

I used cylindrical coordinates to solve this. But I came across a problem.

When I fix z between 0 and 2, and r between 0 and sqrt(2z) I get 16pi/3
{0<z<2; 0<r<sqrt(2z)}

But when I fix r between 0 and 2, and z between 0 and r^2/2 i get 32pi/3
{0<r<2; 0<z<r²/2}
I cannot find error anywhere in the process of calculating. What is the right result?

 

[Clever-Name]

What about the theta dependence?

 

[Bassalisk]

thats 2pi in both cases. 0<theta<2pi

Jacobian = r

 

[Clever-Name]

Oh.. right, I noticed the pi in your results but didn't see the 0<theta<2pi in your setup of the limits. I'll work through it and get back to you.

 

[Bassalisk]

Thanks. Pay attention to my limits. I probably got them wrong in one case.

 

[wbandersonjr]

I am getting (4/3)pi. Also, the graph of it is just an upside down cone correct, so using the volume of a cone formula is also got (4/3)pi, just checking my integration.

 

[Bassalisk]

Well no. Because its x^2+y^2=2z. That is not an upside down cone. Its wide a bit.

Case you are speaking is: x^2+y^2=z^2

 

[wbandersonjr]

Duh. thanks, i see what i did.

 

[Mute]

Here is what you have presumably done in both cases:

$$\int_0^2 dr r \int_0^{2\pi}d\phi \int_0^{r^2/2} dz r^2 = 2\pi \int_0^2 dr r^3 \times \left. z \right|_{0}^{r^2/2} = \pi \int_0^{2} dr~r^5 = \pi \frac{2^6}{6} = \frac{32\pi}{3}$$

$$\int_0^2 dz \int_0^{2\pi} d\phi \int_0^{\sqrt{2z}} dr r^3 = 2\pi \int_0^2 dz \frac{4z^2}{4} = 2\pi \frac{2^3}{3} = \frac{16\pi}{3}$$

I think the trouble is that in the second case, you write $r = \sqrt{2z}$, where you have taken the positive root only. However, the negative root still exists: $(r,\phi) = (-\sqrt{2z},\phi) \equiv (\sqrt{2z},\phi-\pi)$. So, I think you must also add the "negative root" contribution to the second integral:

$$\int_0^2 dz \int_{-\pi}^{\pi} d\phi \int_0^{\sqrt{2z}} dr r^3 = 2\pi \int_0^2 dz \frac{4z^2}{4} = 2\pi \frac{2^3}{3} = \frac{16\pi}{3}$$

Adding them together gives you $32\pi/3$, as in the first case.

 

[Bassalisk]

Here is what you have presumably done in both cases:

$$\int_0^2 dr r \int_0^{2\pi}d\phi \int_0^{r^2/2} dz r^2 = 2\pi \int_0^2 dr r^3 \times \left. z \right|_{0}^{r^2/2} = \pi \int_0^{2} dr~r^5 = \pi \frac{2^6}{6} = \frac{32\pi}{3}$$

$$\int_0^2 dz \int_0^{2\pi} d\phi \int_0^{\sqrt{2z}} dr r^3 = 2\pi \int_0^2 dz \frac{4z^2}{4} = 2\pi \frac{2^3}{3} = \frac{16\pi}{3}$$

I think the trouble is that in the second case, you write $r = \sqrt{2z}$, where you have taken the positive root only. However, the negative root still exists: $(r,\phi) = (-\sqrt{2z},\phi) \equiv (\sqrt{2z},\phi-\pi)$. So, I think you must also add the "negative root" contribution to the second integral:

$$\int_0^2 dz \int_{-\pi}^{\pi} d\phi \int_0^{\sqrt{2z}} dr r^3 = 2\pi \int_0^2 dz \frac{4z^2}{4} = 2\pi \frac{2^3}{3} = \frac{16\pi}{3}$$

Adding them together gives you $32\pi/3$, as in the first case.

Thank you for you reply. Unfortunately that is not the case. You take only positive side because of the condition z>=0;

I solved the problem after half an hour of busting my head. Limits are wrong in the second case, where you get 32pi/3. z goes from r²/2 to 2

{r²/2<z<2}

This makes the result 16pi/3 like in the first case.

I think someone can verify this.

 

[wbandersonjr]

Limits are wrong in the second case, where you get 32pi/3. z goes from r²/2 to 2

{r²/2<z<2}

This makes the result 16pi/3 like in the first case.

.

Yes.

 

[Bassalisk]

Yes.

Thank you