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Math Help - triple integral

  1. #1
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    triple integral

    Evaluate the triple integral \int\int\int_{E} (x + 2y) dV, where E is bounded by the surface y = x^2 and the planes x = z, x = y and z = 0.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by wik_chick88 View Post
    Evaluate the triple integral \int\int\int_{E} (x + 2y) dV, where E is bounded by the surface y = x^2 and the planes x = z, x = y and z = 0.
    I'd like someone to verify this, but I'm pretty sure that this is it.

    Getting the bounds is the hardest part. Just looking at \mathbb{R}^2, the area enclosed by y=x^2 and y=x is the region such that x\in[0,1]. Note that here x\geq x^2.

    Thus, we have

    x: 0\to1
    y: x^2\to x
    z: 0\to x

    Setting up the integral, we have \int_0^1\int_{x^2}^x\int_0^x (x+2y)\,dz\,dy\,dx

    Here's my proposed solution if you want it.

    Spoiler:

    Evaluating the innermost integral gives us (x+2y)z\big|_0^x = x^2+2xy.

    Now the integral is \int_0^1\int_{x^2}^x(x^2+2xy)\,dy\,dx

    Evaluating the next part of the integral gives us (x^2y+xy^2)\big|_{x^2}^x = x^3+x^3-x^4-x^5 = 2x^3-x^4-x^5

    Now the integral is \int_0^1 (2x^3-x^4-x^5)\,dx

    Evaluating this gives us \left(\frac{x^4}{2}-\frac{x^5}{5}-\frac{x^6}{6}\right)\bigg|_0^1 = \frac{1}{2}-\frac{1}{5}-\frac{1}{6} = \frac{2}{15}

    So I think \frac{2}{15} is the answer.
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