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I am wondering if there is a difference between the answers of these when getting the volume of solid of revolution:

Region bounded by $y = x$, $x = 0$ and $y = 15$, rotated about the $x-axis$

Region bounded by $y = x$, $x = 0$ and $y = 15$, rotated about the $y-axis$

I tried the first one which is rotated about the x-axis and I got $2250π$.

For the second one I am confused on how to solve, is it possible that I use any of disk, washer, or shell method?

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  • $\begingroup$ What did you get? What is $2250\pi$? Is it volume? $\endgroup$
    – VIVID
    Apr 29 '21 at 9:02
  • $\begingroup$ @VIVID Sorry, I forgot to indicate it. Yes, volume :) $\endgroup$ Apr 29 '21 at 9:05
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    $\begingroup$ When you rotate around y-axis, you get a cone with height $h = 15$ and radius $r = 15$ so the volume will be $1125 \pi$. When you rotate around x-axis, it is the cone of same volume cut out from a cylinder of height $h = 15$ and radius $15$. So you get $2250 \pi$. $\endgroup$
    – Math Lover
    Apr 29 '21 at 9:20
  • $\begingroup$ @VIVID Thank you, can I use any of the methods, disk, washer, or shell? $\endgroup$ Apr 29 '21 at 9:22
  • $\begingroup$ Can you describe those methods? I agree with @MathLover and so does my answer. Try this! $\endgroup$
    – VIVID
    Apr 29 '21 at 9:23
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No objects are not the same.

  • Your region is the following triangle:

enter image description here


  • Rotation around the $x$-axis:

enter image description here


  • Rotation around the $y$-axis:

enter image description here


However, you should be able to see the relation between their volumes.

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Volume of the figure produced by rotation is equal to $V = 2 *S * π * r$

Where S is area of cross-section, r is distance of the center of mass from rotation axis.

For x it is 10, for y it is 5, see pic.

So we get for the first case $2 *1/2 * 15^2 * 10 = 2250 π$ In second case $2 *1/2 * 15^2 * 5 = 1125 π$

plot of the area

enter image description here

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