Find the volume of the solid that results when the region enclosed by y=x^1/2, y=0, and x=121 is revolved about the line y=121

What we have here basically is two revolutions, so we can subtract the volume of the inner from the outer to get the final answer. To find the volume of each, we just integrate the area of the circle (pi * r ^ 2) over our interval (in this case from x=0, since sqrt(x) is not real where x < 0, to the given x=121). So with subtracting/simplifying: \[V = \pi \int\limits\limits\limits_{0}^{121}R_{outer}(x)^2 - R_{inner}(x)^2dx \] Where R(x) is the radius as a function of x. The radius of the outer circle is going to be (121 - 0), or 121, the distance between y=0 and the axis of revolution. The radius of the inner revolution, then, is going to be (121 - sqrt(x)). If we plug these into our equation, we get: \[V = \pi \int\limits\limits\limits\limits_{0}^{121}121^2 - (121 - \sqrt{x})^2dx\] So V equals about 651610.8823.

Thank you so much. : ) I see where I went wrong in my work. For some reason I put my limits from 0 to 11 instead of 0 to 121. Thanks again.

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