Why doesn't this method work for getting volume of a sphere? And how to find volume of a spherical cap?

190 Views Asked by At

As far as I dig into higher mathematics, I found a problem myself for getting volume of a spherical cap, as I was doing some experiments with Archimedes Principle.

spherical capspherical cap handwritten

So what I did was take elemental mass in terms of angle θ taking center of sphere as origin.

So, lets see what I did and what i was getting

$$r = R\cos θ$$ (vertically-opposite angle, see Z looking in figure)

Now volume of small elemental mass $$dV = πr^2\, dl$$ (where $dl$ is thickness) $$dl = R\,dθ$$ Putting all together $$dV = π(R\cos θ)^2(R\,dθ)$$ $$dV = πR^3\cos^2θ\,dθ$$

Now total volume of the sphere if i want to calculate to check this formula is working correct, $$V = \int_{-π/2}^{π/2} πR^3 \cos^2θ\,dθ$$ $$V = πR^3 \int_{-π/2}^{π/2} \cos^2θ\,dθ$$ $$V = πR^3 \int_{-π/2}^{π/2} \frac{1+\cos2θ}{2}\,dθ$$ $$V = πR^3 \left[ \frac{θ}{2} + \frac{\sin 2θ}{4} \right]_{-π/2}^{π/2}$$ $$V = \frac{π^2R^3}{2}$$

But this contradicts the fact that Volume of sphere is $\frac{4}{3}πR^3$

And in fact, the surface area of the sphere is taken out of this method to, instead taking $πr^2\,dl$ we use $2πr\,dl$ to get elemental area and then integrate from 0 to $\frac{π}{2}$ and then twice that. But doing that will not gonna help in getting volume of a spherical cap, so I took elemental volume.

What is done wrong in this method, any help would be appreciated. :)

2

There are 2 best solutions below

8
On BEST ANSWER

Yes, $r=R \cos \theta$ is OK. But also $z=R \sin \theta$ so $dz=R \cos \theta d \theta.$ Then you end up integrating $\pi (R \cos \theta)^2 \cdot (R \cos \theta d \theta)$ i.e. $\pi R^3 (\cos \theta)^3,$ and the cube of cosine integrates to $4/3$ when done $-\pi/2$ to $\pi/2.$

2
On

You have a mistake in determining the thickness. By looking at this picture, you can see that the thickess $h$ isn't the same as the length of an arc $Rd\theta$, and you need an additional factor of $\cos\theta$, which makes your integral equal to $\frac43\pi R^3$ as it should be.

enter image description here