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?
On
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
No objects are not the same.
However, you should be able to see the relation between their volumes.