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Math Help - Volume between surfaces

  1. #1
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    Volume between surfaces

    I have a couple problems about calculating volumes between surfaces and I don't feel 100% confident about my answers. Is there an easy way to check my answers? I'm not very wolfram-savvy, as I failed to even figure out how to plot all the surfaces in one graph. Is there a way to use wolfram or some other tool to check my answers? Failing that, I could just type up the integrals but I figured I could save everyones time by asking the above. Any suggestions?
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    Re: Volume between surfaces

    Unfortunately a computer can not think, so it has no way to determine if what you are putting in is the right model for your situation.

    The only way you're going to know if you have set up the integrals correctly is to post your question and attempt at solution here for us to check.
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    Re: Volume between surfaces

    I'm just saying if you can graph the surfaces on a computer you can probably calculate volume too. In any case, these are the problems.


    1.Volume bounded by x=0, x + y + z = 10 , z=0 and z = 4-(y2). I set up the integral as (-2,2) (0,4-(y2) ) (0,10-y-z) dx dz dy. My volume was 24.
    2.Volume bounded by y = z2, y=16, x+y+z=3, x+y+z=30. I got (-4,4) (z^2 ,16) (3-y-z,30-y-z) dx dy dz. My volume was 2304
    3.Volume bounded by z = x2 + y2, z=-4, x+y=1, x=0, y=0. I got (0,1) (0,1-y) (-4, x2 + y2 ) dz dx dy. My volume was 13/6.
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    Re: Volume between surfaces

    Quote Originally Posted by tleaf555 View Post
    I'm just saying if you can graph the surfaces on a computer you can probably calculate volume too. In any case, these are the problems.


    1.Volume bounded by x=0, x + y + z = 10 , z=0 and z = 4-(y2). I set up the integral as (-2,2) (0,4-(y2) ) (0,10-y-z) dx dz dy. My volume was 24.
    2.Volume bounded by y = z2, y=16, x+y+z=3, x+y+z=30. I got (-4,4) (z^2 ,16) (3-y-z,30-y-z) dx dy dz. My volume was 2304
    3.Volume bounded by z = x2 + y2, z=-4, x+y=1, x=0, y=0. I got (0,1) (0,1-y) (-4, x2 + y2 ) dz dx dy. My volume was 13/6.
    I agree with the setup of your first triple integral, so

    \displaystyle \begin{align*} V &= \int_{-2}^2{\int_0^{4 - y^2}{\int_0^{10 - y - z}{\,dx}\,dz}\,dy} \\ &= \int_{-2}^2{\int_0^{4 - y^2}{\left[x\right]_0^{10 - y - z}\,dz}\,dy} \\ &= \int_{-2}^2{\int_0^{4 - y^2}{10 - y - z\,dz}\,dy} \\ &= \int_{-2}^2{\left[10z - yz - \frac{1}{2}z^2\right]_0^{4 - y^2}\,dy} \\ &= \int_{-2}^2{10\left(4 - y^2\right) - y\left(4 - y^2\right) - \frac{1}{2}\left(4 - y^2\right)^2\,dy} \\ &= \int_{-2}^2{40 - 10y^2 - 4y + y^3 - 8 + 4y^2 - \frac{1}{2}y^4\,dy} \\ &= \int_{-2}^2{-\frac{1}{2}y^4 + y^3 - 6y^2 - 4y + 32 \,dy} \\ &= \left[-\frac{1}{10}y^5 + \frac{1}{4}y^4 - 2y^3 - 2y^2 + 32y\right]_{-2}^2 \\ &= \left[ -\frac{1}{10}(2)^5 + \frac{1}{4}(2)^4 - 2(2)^3 - 2(2)^2 + 32(2)\right] - \left[-\frac{1}{10}(-2)^5 + \frac{1}{4}(-2)^4 - 2(-2)^3 - 2(-2)^2 + 32(-2)\right] \\ &= \left(-\frac{16}{5} + 4 - 16 - 8 + 64\right) - \left(\frac{16}{5} + 4 + 16 - 8 - 64\right) \\ &= -\frac{32}{5} + 96 \\ &= -\frac{32}{5} + \frac{480}{5} \\ &= \frac{448}{5} \end{align*}
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