Area and Volume Formulas — Question 10

Question 10 - Solution

We derive the formula using the method of cross-sectional disks.

Place the sphere of radius $R$ so that it is centered at the origin. Its equation is $$x^2+y^2+z^2=R^2.$$

Fix a value of $x$ with $-R\le x\le R$. A cross-section perpendicular to the $x$-axis is a circle whose radius depends on $x$.

Solving the equation of the sphere for $y^2+z^2$ gives $$y^2+z^2=R^2-x^2.$$

Thus, the radius of the cross-sectional disk at position $x$ is $$r(x)=\sqrt{R^2-x^2}.$$

The area of this circular cross-section is $$A(x)=\pi[r(x)]^2=\pi(R^2-x^2).$$

An infinitesimally thin slice of thickness $dx$ has volume $$dV=A(x)\,dx=\pi(R^2-x^2)\,dx.$$

Integrate from $x=-R$ to $x=R$ to obtain the total volume: $$V=\int_{-R}^{R}\pi(R^2-x^2)\,dx.$$

Factor out $\pi$: $$V=\pi\int_{-R}^{R}(R^2-x^2)\,dx.$$

Evaluate the integral: $$\int_{-R}^{R}R^2\,dx=2R^3, \qquad \int_{-R}^{R}x^2\,dx=\frac{2R^3}{3}.$$

Thus, $$\int_{-R}^{R}(R^2-x^2)\,dx = 2R^3-\frac{2R^3}{3} = \frac{4R^3}{3}.$$

Substitute back: $$V=\pi\cdot\frac{4R^3}{3} =\frac{4}{3}\pi R^3.$$

$$\boxed{V=\frac{4}{3}\pi R^3}$$