Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y=sqrt(25-x^2), y=0, x=2; about the x-axis

Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
y=sqrt(25-x^2), y=0, x=2; about the x-axis

anonymous · Answer

I am having trouble setting up the integral.

kainui · Answer

Show me what you have and I'll help you fix it.

anonymous · Answer

$$V=\int\limits_{-5}^{5}\pi[(\sqrt{25-x^2})^2-(2)^2]dx$$

anonymous · Answer

But I'm not sure because I know I that is used for the inner radius and outer radius. The part that is confusing me is the x=2.

kainui · Answer

Hmm. Well the 2^2 part subtracted off is where you've gone wrong. Really your upper bound should be x=2 not x=5. The -2^2 part shouldn't be there at all. |dw:1397181538845:dw|

anonymous · Answer

Oh okay. I see where you got the upper bound from. So should it be$$V=\int\limits_{-5}^{2}p[i(\sqrt{25-x^2})]^2ds$$

anonymous · Answer

that should be pi where it says p[i

kainui · Answer

Yes exactly. See, the integral is adding up the infinite number of infinitely small volumes of circles from -5 to 2 on the x axis. Each circle has a radius of the function at every point x, and so you have the area multiplied by dx, which is infinitesimally small.

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anonymous · Answer

So for the answer I got 392pi/3.

kainui · Answer

I did not calculate it, but that looks about right to me.

anonymous · Answer

Okay thank you!