I need help on 2 calculus questions, Volume of A Solid of Revolution?

When you revolve a curve about some axis, you produce a solid which has a circular body with some height. At different height, the circular body has different radii. Thus, the magnitude of the radii is related to the heights. That is, r can be a function of h or h can be a function of r.

2πr is the circumference of the particular circle.

To find the volume, we write

∫∫ 2πr drdh or ∫∫ 2πr dhdr

, because change the sequence of the integration won't change the answer.

∫∫ 2πr drdh = ∫ πr² dh is so called disk method.

∫∫ 2πr dhdr = ∫ 2πrh dr is so called shell method.

Using disk method, we need to express r as a function of h so we can integrate.

Using shell method, we then need to express h as a function of r to integrate.

Therefore, which method to choose depends on if r can be a function of h or the other way around. Sometimes, both ways are fine; sometimes only one would work.

Now regarding rotating about x axis or y axis.

Rotating about x axis makes y become r and x as h;

Rotating about y axis makes x become r and y as h.

Other than that, nothing new.

One more note about the area:

∫∫∫rdrdθdh is the volume expression in cylindrical coordinates.

Within the scope we are discussing here, we integrate wrt θ first. That is, ∫∫∫rdθdrdh = ∫∫∫ 2πrdrdh

Yes, you're correct. If you are slicing the volume horizontally into disks, then the volume of each disk is pi*r^2*dh. r will depend on h.

2pi*r*h is the surface area, not teh volume.