Finding a double integral that represents the given solid

[ksdhart]

I'm having difficulties with a problem from my Calculus IV course. The problem states:

In exercises 47-56, use polar coordinates to find an iterated integral that represents the volume of the solid described and then find the volume of the solid.

52) The region bounded above by the unit sphere centered at the origin, and bounded below by the plane z = h, where 0 <= h <= 1

My book has exactly one example even close to problem, and we didn't go over any other examples in class, so I'm a bit confused. This is the double integral I came up with, but I think it's wrong.

\(\displaystyle \int _0^{2\pi }\int _0^1\:\:r\cdot \left(\sqrt{1-r^2}-h\right)drd\theta \)

Specifically, the fact that h is bounded between 0 and 1 is what's giving me grief. What does that mean in the context of this problem, and how can I account for it in my integral?

 

[Prove It]

I'm having difficulties with a problem from my Calculus IV course. The problem states:



My book has exactly one example even close to problem, and we didn't go over any other examples in class, so I'm a bit confused. This is the double integral I came up with, but I think it's wrong.

\(\displaystyle \int _0^{2\pi }\int _0^1\:\:r\cdot \left(\sqrt{1-r^2}-h\right)drd\theta \)

Specifically, the fact that h is bounded between 0 and 1 is what's giving me grief. What does that mean in the context of this problem, and how can I account for it in my integral?

All that \(\displaystyle \displaystyle \begin{align*} 0 \leq h \leq 1 \end{align*}\) means is that the plane is parallel to the x-y plane and won't go past the sphere (as the sphere has a maximum height of 1 - can you see why?)

Bounded ABOVE by the unit sphere centred at the origin => your upper function is \(\displaystyle \displaystyle \begin{align*} x^2 + y^2 + z^2 = 1 \implies z = \sqrt{ 1 - \left( x^2 + y^2 \right) } = \sqrt{1 - r^2} \end{align*}\)

Bounded BELOW by the plane \(\displaystyle \displaystyle \begin{align*} z = h \end{align*}\) where \(\displaystyle \displaystyle \begin{align*} 0 \leq h \leq 1 \end{align*}\).

So your integrand in polars is \(\displaystyle \displaystyle \begin{align*} \sqrt{ 1 - r^2 } - h \end{align*}\).

Now think about every cross section parallel to the x-y plane. They are circles. The largest circle will be at the bottom of your region, so where \(\displaystyle \displaystyle \begin{align*} z = \sqrt{1 - r^2} = h \implies r = 1 - h^2 \end{align*}\) and obviously the smallest circle will be right at the top of your sphere, where \(\displaystyle \displaystyle \begin{align*} r = 0 \end{align*}\). Since every circle is complete that means \(\displaystyle \displaystyle \begin{align*} 0 \leq \theta \leq 2\,\pi \end{align*}\). So that means your double integral is

\(\displaystyle \displaystyle \begin{align*} V = \int_0^{2\,\pi}{ \int_0^{ \sqrt{1 - h^2} }{ \left( \sqrt{1 - r^2} - h \right) \, r\,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*}\)

 

[ksdhart]

Oh, wow. Your explanation makes so much sense to me. I wish the people who wrote my textbook explained it like that. As I experience more of this textbook, I'm getting the impression that it wasn't really written for students trying to learn the material for the first time. Rather, it feels like it's meant to be a refresher for someone who already knows the material. In any case, thank you so much for helping me out.