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

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?

• What did you get? What is $2250\pi$? Is it volume? Apr 29 '21 at 9:02
• @VIVID Sorry, I forgot to indicate it. Yes, volume :) Apr 29 '21 at 9:05
• 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 $2250 \pi$. Apr 29 '21 at 9:20
• @VIVID Thank you, can I use any of the methods, disk, washer, or shell? Apr 29 '21 at 9:22
• Can you describe those methods? I agree with @MathLover and so does my answer. Try this! Apr 29 '21 at 9:23

No objects are not the same.

• Your region is the following triangle:

• Rotation around the $$x$$-axis:

• Rotation around the $$y$$-axis:

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

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