# Volume sphere and integration - slant height

Consider the ball in $\mathbb{R}^3$ of radius $R$ and two planes of height $z$ and $z+dz$ which intersect the sphere in two circles of radius $r(z)$ and $r(z+dz)$.

The volume $dV$ of the piece of the ball between this two planes is approached by the volume of the cylinder with basis the circle of radius $r(z)$ and of height $dz$, that is, $$dV(z) = \pi r(z)^2 dz$$

Then the volume of the ball is given by $V = \int_{-R}^RdV(z)$.

Now, take $\theta$ the angle locating the circle of radius $r(z)$ from the center of the ball so that $Rd\theta$ is the arc length of the slant height of the element of volume $dV$.

Why the following expression lead to an other expression for the volume $V$ when integrating on $\theta \in [-\pi/2,\pi/22]$ ? $$dV(\theta) = \pi r(z)^2 Rd\theta$$

EDIT: Instead of approximate the volume of the slice by the volume of a cylinder (which gives the right formula), I follow the answer of @amd and I approximate the volume of the slice of the sphere by the volume of the a frostum of a cone (intuitively, it is more precise). The latter is equal to: $$dV_{slice}(z) = \pi/3(r(z)^2+r(z)r(z+dz)+r(z+dz)^2)\,dz$$ and the volume of the sphere is $$V_S = \int_{-R}^R dV_{slice}(z) = \ldots$$ but I can't handle the computations to retrieve the right formula.

Does this method fail ?

• When integrating a sphere, intuitively, you have to be sure that the infinitely thin parts you are adding up to arrive at the volume, are in a sense, of "uniform thickness". So if you integrate spheres out from the centre you get the volume of a ball. But if you integrate circumferences around through $2\pi$ it doesn't work because the equator moves faster than the poles. – samerivertwice Apr 17 '18 at 13:50

Since you want to integrate on $\left[-\frac\pi2,\frac\pi2\right],$ presumably you are measuring $\theta$ so that it is $0$ when the circle has radius $R$ (at the equator of the sphere).
In any case, the formula you have shown is not correct. If you are measuring $\theta$ as I think you are, the formula should be $$dV(\theta) = \pi (r(z))^2 R\cos\theta\, d\theta.$$
This formula can be integrated to get the volume of the sphere: $$\int_{-\pi/2}^{\pi/2} dV(\theta) = \int_{-\pi/2}^{\pi/2} \pi (r(z))^2 R\cos\theta\, d\theta = \int_{-\pi/2}^{\pi/2} \pi (R^2 - (R\sin\theta)^2) R\cos\theta\, d\theta.$$
The reason you need the factor of $\cos \theta$ is that $R\,d\theta$ is measured in a direction that is at an angle $\theta$ from vertical; that's why it is called a slant height. The volume (in the limit) of the slice of the sphere between the two disks at $z = R\sin\theta$ and $z = R\sin(\theta + d\theta)$ is measured by multiplying the area of the base disk by the perpendicular (not slant) distance between the disks.
The perpendicular distance between the disks is one leg of a right triangle where $R\,d\theta$ is the hypotenuse. The angle between that leg and the hypotenuse is $\theta$, so the leg is $R \cos\theta\,d\theta.$