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## anonymous one year ago Consider the region under the curve y = 1/x, above the x axis, on the intervalbum 1 <= x < infinity. Calculate the volume of the solid obtained by revolving the region about the x axis. Delete Cancel Submit

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1. Astrophysics
• one year ago
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@ganeshie8

2. Astrophysics
• one year ago
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|dw:1436287913646:dw|

3. Astrophysics
• one year ago
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$V = \int\limits_{a}^{b} A(x) dx~~~~~~\text{where} ~ A(x) = \pi r ^2$

4. Astrophysics
• one year ago
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So I guess you can use the disk method, but it's been a while since I've done such problems, the radius for disk is f(x)

5. Astrophysics
• one year ago
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So $V = \int\limits_{a}^{b} \pi f(x)^2 dx$

6. Astrophysics
• one year ago
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Now just plug in the intervals and values and you're good to go!

7. anonymous
• one year ago
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Can you show me

8. Astrophysics
• one year ago
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$V = \int\limits_{1}^{\infty} \pi \left( \frac{ 1 }{ x } \right)^2 dx$

9. Astrophysics
• one year ago
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Go ahead and try to integrate now

10. anonymous
• one year ago
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