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Math Help - Triple Integrals

  1. #1
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    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.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dude15129 View Post
    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<y<\sqrt{1-x^2}

    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
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  3. #3
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    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.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dude15129 View Post
    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...

    Quote Originally Posted by Chris L T521 View Post
    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
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