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I've been working on with the area of the region in my calc II class, and now have to deal with the volume.

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. $$y = 7\sqrt{49 − x^2}, y = 0, x = 2, x = 3$$ about the x-axis

I was able to draw the graph, but can't set up the integral. Any help would be appreciated. Thanks!

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Try using the disk method.

$$ \pi \int_2^3 \left(7\sqrt{49 - x^2}\right)^2 \ dx $$

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The region you are rotating about the x-axis can be visualized as follows: enter image description here

If you recall the disk method, the way we visualize finding the volume of the solid of revolution around the x-axis, is, we are essentially adding up infinitely thin disks of volume $dV$ for each small width $dx$ at each $x$ (a way to think of the integral). A disk at position $x$ will have radius equal to the distance from the x-axis to the curve, in this case $7\sqrt{49-x^2}$, and will have width $dx$, an infinitely small piece of length on the x-axis. Therefore $$ dV = \pi r^2 h = \pi (7\sqrt{49-x^2})^2 \, dx = 49\pi(49-x^2) \, dx $$ and so \begin{align} V &= 49\pi \int_2^3 (49-x^2) \, dx \\ &= 49\pi \left[ 49-x^2 \right]_2^3 \\ &= 49\pi \left[ 49x-\frac{x^3}{3} \right]_2^3 \\ &= 49\pi \left[ 49-\frac{19}{3} \right] \\ &= 49\pi \cdot \frac{128}{3} \\ &= \frac{6272\pi}{3} \end{align}

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