# Please help with this volume of solid of rotation question

Find the volumes of the solids generated by revolving the region bounded by the graphs of the equations about the y-axis.

$$y = \sqrt(x), y=0, x=6$$

The reason this is confusing to me is because it's asking me to take an area that is bounded by the x axis, a function, and another line, and rotate it about the y axis. If the area were bounded by the y axis instead of the x axis, this would make much more sense to me.

If I want to revolve the function about the y axis, I should make it in terms of y: $$y=\sqrt{x} , x=y^2$$ So to integrate with respect to y using $$0$$ to $$6$$ on the x-axis as my bounds, I have to convert my bounds to be on the y-axis, and they become $$0$$ and $$\sqrt{6}$$. Then, I can try to integrate. The problem is that if I just integrate the function normally with respect to y, I'm finding the area bound by the function and the y axis: $$\int_0^\sqrt{6}y^2dy$$

So, with respect to y, I tried treating it like the area between the function $$x=6$$ and $$y^2$$: $$\int_0^\sqrt{6} 6-y^2 dx$$

Then, using the disk method formula, the volume would be: $$π\int_0^\sqrt{6} (6-y^2)^2dy$$ Evaluate that and you get $$\pi(\frac{96\sqrt{6}}{5})$$, which my online homework's autochecker said was wrong. Could someone tell me my error in reasoning here? Thank you.

When the region you are trying to rotate does not touch the axis of rotation, the "disks" you get have circular holes in the middle.

Some people use the term washer method for this technique. where the "washer" is a disk with a hole in the middle, like the washers that one might use when assembling some mechanical object with nuts and bolts.

There are some explanations here and here. Note that the second link calls this the method of disks or the method of rings, but the formula for rotation around the $$y$$ axis is the same no matter what you call the method:

$$\int_a^b A(y)\,dy = \int_a^b \pi ((\text{outer radius})^2 - (\text{inner radius})^2)\, dy.$$

This is because $$A(y)$$ is the area of a disk-with-a-hole (also called washer or ring or annulus) with an outer radius and an inner radius. If it were a disk with the given outer radius and no hole, its area would be $$\pi (\text{outer radius})^2,$$ but the hole removes $$\pi (\text{inner radius})^2$$ from the area.

What you did was $$\int_a^b \pi ((\text{outer radius}) - (\text{inner radius}))^2\, dy$$

which is effectively the area of a smaller disk without a hole. You get a smaller result that way.