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Math Help - Help with work problems

  1. #1
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    Help with work problems

    Let R be the region bounded by the curves y=x^2, y=sin(xpi/2), x=0, and x=1.
    Find the volume of the solid obtained by rotating R around the y-axis.
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  2. #2
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    Note that: \sin \left(\frac{\pi x}{2}\right) \geq x^{2} \quad 0 \leq x \leq 1

    A graph of this would be helpful in visualizing what's going on.

    By method of shells: V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx

    x = the radius of the shell; \sin\left(\frac{\pi}{2}x\right) - x^{2} = represents the height of te shell
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  3. #3
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    thanks

    how would i solve that integral? it looks complicated
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  4. #4
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    It's not too bad:

    V = 2\pi \int_{0}^{1} x \left(\sin\left(\frac{\pi}{2}x\right) - x^{2}\right)dx
    V = 2\pi \int_{0}^{1} \left[ x\sin \left(\frac{\pi}{2} x\right) - x^{3}\right] dx

    You can integrate term by term (since: \int \left(f(x) + g(x)\right) dx = \int f(x) dx + \int g(x) dx) in which case the 2nd term should be rather easy.

    The first one, i.e. \int_{0}^{1} x\sin \left(\frac{\pi}{2}x\right) dx requires integration by parts:
    \begin{array}{lll} u = x & \quad & dv = \sin \left(\frac{\pi}{2}x\right) \\ du = dx & \quad & v = -\frac{2}{\pi} \cos \left(\frac{\pi}{2}x\right) \end{array}

    Sub it into the integration by parts formula and you should be on your way.
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