# Math Help - Integral

1. ## Integral

I have the problem mostly done but this last part of the integration is getting to me. So we originally had a triple integral that I had to integrate with respect to z. The limits were from -(sqrt(13-4x^2))/3 and to the positive of that. So we ended up getting (z^3)/3 to plug in the limits to and, after simplifying, we have (2(13-4x^2)^3/2)/81 and I have to take that with respect to x. I simplified it a bit further and now am down to 26/81 on the outside to be multiplied in and (-8x^2 (sqrt(13-4x^2)) needing to be integrated. I was thinking substitution to get the sqrt to be u but then we would still be left with an x when we pull out the -8xdx. I was also thinking of doing it by part but I'm not sure if that would work either. Bah.

2. Originally Posted by Centara
I have the problem mostly done but this last part of the integration is getting to me. So we originally had a triple integral that I had to integrate with respect to z. The limits were from -(sqrt(13-4x^2))/3 and to the positive of that. So we ended up getting (z^3)/3 to plug in the limits to and, after simplifying, we have (2(13-4x^2)^3/2)/81 and I have to take that with respect to x. I simplified it a bit further and now am down to 26/81 on the outside to be multiplied in and (-8x^2 (sqrt(13-4x^2)) needing to be integrated. I was thinking substitution to get the sqrt to be u but then we would still be left with an x when we pull out the -8xdx. I was also thinking of doing it by part but I'm not sure if that would work either. Bah.
Let $x = \frac{\sqrt{13}}{2}sin(\theta)$.

-Dan

3. Confused. Again. How did you get that and how would it help? This is why I'm dropping my math major once this class is over, lol. i can't see things like other people do.

4. Hello,

-(sqrt(13-4x^2))/3
The trick is to get something like a²-a²z², then to substitute z by sin(x) or cos(x).
Hence $\sqrt{a^2-(az)^2}=\sqrt{a^2(1-z^2)}=|a| \sqrt{1-sin^2(x)}=|a| \cdot |cos(x)|$, which is easy to integrate. That's why we make this transformation

~~~~~~~~~~~~

$13-4x^2=(\sqrt{13})^2-\frac{13}{13} \cdot 2^2 x^2$

$=(\sqrt{13})^2-(\sqrt{13})^2 \cdot \frac{2^2}{13} x^2$

$a=\sqrt{13}$

$\dots =a^2-a^2 \cdot \left(\sqrt{\frac{2^2}{13}} \cdot x \right)^2$

$=a^2-a^2 \cdot \left(\frac{2}{\sqrt{13}} \cdot x \right)^2$

-----> $\sin(\theta)=\frac{2}{\sqrt{13}} \cdot x$

---------> $x=\frac{\sqrt{13}}{2} \cdot \sin(\theta)$

5. Okay, so I honestly don't think my professor would give us something like that because that gives me a headache, lol. I'm wondering if maybe my limits are wrong or I messed up along the way. The question is: Find the volume bounded by the paraboloids y=9x^2 + 4z^2 and y=13-4x^2 - 9z^2

I got the limits of y, the first of the equations being the lower and the second as the upper. Then the one's I found for z and then x being from -1 to 1. The equation i integrated was just 1 then dydzdx. Anyone see a problem with any of that? Or is it just a majorly hard problem?

6. Originally Posted by Centara
Okay, so I honestly don't think my professor would give us something like that because that gives me a headache, lol. I'm wondering if maybe my limits are wrong or I messed up along the way. The question is: Find the volume bounded by the paraboloids y=9x^2 + 4z^2 and y=13-4x^2 - 9z^2

I got the limits of y, the first of the equations being the lower and the second as the upper. Then the one's I found for z and then x being from -1 to 1. The equation i integrated was just 1 then dydzdx. Anyone see a problem with any of that? Or is it just a majorly hard problem?
The paraboloids intersect on the circle $x^2 + z^2 = 1$. So you have to integrate $(13 - 4x^2 - 9z^2) - (9x^2 + 4z^2)$ in the xz-plane over the region bounded by this circle. Switching to polar coordinates would probably be a smart thing to do.