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Math Help - Volume of a solid

  1. #1
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    Volume of a solid

    My friend and I ran into another problem that we were not able to solve. If anyone can gives us a hint, we'd really appreciate it.

    A solid is generated by rotating the curve y=f(x) where 0\leqslant x \leqslant c about the x-axis, and the resulting volume is c^2 + c. We want to find the function f(x).
    We have the volume using disk method V=\int_0^c \pi r^2 dx=c^2+c. But r=y=f(x), and I don't know how to change the variable of the expression inside the integral so that I can integrate it.
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  2. #2
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    Quote Originally Posted by jackie View Post
    My friend and I ran into another problem that we were not able to solve. If anyone can gives us a hint, we'd really appreciate it.

    A solid is generated by rotating the curve y=f(x) where 0\leqslant x \leqslant c about the x-axis, and the resulting volume is c^2 + c. We want to find the function f(x).
    We have the volume using disk method V=\int_0^c \pi r^2 dx=c^2+c. But r=y=f(x), and I don't know how to change the variable of the expression inside the integral so that I can integrate it.
    1. Rotation about the x-axis is calculated by:

    V = \int\left( \pi \cdot (f(x))^2\right) dx

    That means the r in your equation correspond to the function f.

    2. Now calculate backwards:

    V = \int_0^c \left( \pi \cdot (f(x))^2\right) dx = c^2+c

    \int_0^c \left( (f(x))^2\right) dx = \dfrac1{\pi} (c^2+c)

    Change variables and differentiate both sides:

    (f(x))^2 = \dfrac2{\pi} x + \dfrac1{\pi}~\implies~f(x) = \sqrt{\dfrac2{\pi} x + \dfrac1{\pi}}
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