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Math Help - Double integrals to find the volume of a solid

  1. #1
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    Double integrals to find the volume of a solid

    Example 3. Find the volume of the solid enclosed by surface y^2 = x and by planes z = 0 and x + z = 1.

    I am stuck on what to put for limits, I keep getting -infinity for my answer.

    -x+1 dxdy I get 0 to -infinity for the first integral and -sqrt y to + sqrt y for the second integral.

    Thanks in advance.
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  2. #2
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    Solve x+z=1 for z and get z=1-x.

    \int_{-1}^{1}\int_{y^{2}}^{1}(1-x)dxdy
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  3. #3
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    y^2= x is a "parabolic cylinder" opening in the positive x direction. It is bounded below by z= 0 and above by z= 1- x. Fortunately those two intersect in the line x= 1, bounding the cylinder on the right. In the x,y plane, the figure is bounded by y^2= x on the left, top, and bottom, and by x= 1 on the right.

    You could take x from 0 to 1 and, for each x, y= -\sqrt{x} to y= \sqrt{x}. For each x,y, z goes from 0 up to 1- x so the integrand is 1-x-0= 1- x.

    The volume is given by \int_{x= 0}^1\int_{y= -\sqrt{x}}^{\sqrt{x}} (1- x) dy dx.

    But, of course, it is easier to reverse the order of integration. The lowest y value on the entire figure is -1 at (1,-1) and the highest is 1 at (1,1), both where the parabola intersects the line x= 1. For each y, then x ranges from the parabola x= y^2 to the line x= 1.

    The volume is given by \int_{y=-1}^1\int_{x= y^2}^1 (1- x) dx dy.
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