# Thread: Triple Integrals

1. ## Triple Integrals

I am having trouble with these triple integrals. I understand them when it already tells me what to integrate from...but I have trouble figuring it out with just the given bounded area. Here is one of the ones I am stuck on.

∫∫∫ xdV, where D is bounded by the paraboloid z=x^+y^2, and the plane z=1

I really have no clue how to set it up.

2. Originally Posted by dude15129
I am having trouble with these triple integrals. I understand them when it already tells me what to integrate from...but I have trouble figuring it out with just the given bounded area. Here is one of the ones I am stuck on.

∫∫∫ xdV, where D is bounded by the paraboloid z=x^+y^2, and the plane z=1

I really have no clue how to set it up.
You're given your z limits:

$1\leq z\leq x^2+y^2$

Looking at the cross section of the paraboloid at $z=1$, we see that we have a circle of radius 1. This means that $0\leq x\leq 1$ and $0

Thus, our triple integral is $\iiint\limits_Gx\,dV=4\int_0^1\int_0^{\sqrt{1-x^2}}\int_1^{x^2+y^2}x\,dz\,dy\,dx$

The reason why its multiplied by for is because of how I defined my x and y boundaries. It took into consideration only 1/4th of the region $R$, so in order to get the correct value after integrating it, you need to quadruple the value...

Does this make a little more sense?

--Chris

3. ok...so when you do this....you get:

4∫∫ (x^3 + xy^2 - x) dydx

4∫ (x^3y + 1/3 xy^3 -xy) dx *that is from 0 to √(1-x^2)

4∫ (x^3(√(1-x^2)) +1/3x(1-x^2)^3 - x√(1-x^2))dx

This is just a messy integral...if I did it correctly.

4. Originally Posted by dude15129
ok...so when you do this....you get:

4∫∫ (x^3 + xy^2 - x) dydx

4∫ (x^3y + 1/3 xy^3 -xy) dx *that is from 0 to √(1-x^2)

4∫ (x^3(√(1-x^2)) +1/3x(1-x^2)^3 - x√(1-x^2))dx

This is just a messy integral...if I did it correctly.
It does look messy...

Originally Posted by Chris L T521
Thus, our triple integral is $\iiint\limits_Gx\,dV=4\int_0^1\int_0^{\sqrt{1-x^2}}\int_1^{x^2+y^2}x\,dz\,dy\,dx$
I just thought of an easier way to evaluate this. Convert the integral from rectangular to cylindrical coordinates:

$4\int_0^1\int_0^{\sqrt{1-x^2}}\int_1^{x^2+y^2}x\,dz\,dy\,dx=\int_0^{2\pi}\i nt_0^1\int_1^{r^2}r\cos\vartheta\cdot r\,dz\,dr\,d\vartheta$ $=\int_0^{2\pi}\int_0^1\int_1^{r^2}r^2\cos\vartheta \,dz\,dr\,d\vartheta$

I used the fact that the region is a circle of radius 1. This means that the radius is bounded from $0\leq r\leq 1$ and the angle is defined from $0\leq\vartheta\leq2\pi$ (to go around the whole circle). Then use the identities $x^2+y^2=r^2$ and $x=r\cos\vartheta$ to convert the rest of the integral over to cylindrical coordinates.

Does this conversion process make sense? Now it should be a lot easier to integrate!!

--Chris