Area and Volume Formulas — Question 7

Question 7

Prove that the area of a circle of radius $R$ is $$A=\pi R^2.$$

Question 7 - Solution

We derive the area formula using integration.

Consider a circle of radius $R$ centered at the origin. Its equation is $$x^2+y^2=R^2.$$

Solving for $y$ gives the upper semicircle: $$y=\sqrt{R^2-x^2}.$$

The area of the entire circle is twice the area of the upper semicircle: $$A=2\int_{-R}^{R}\sqrt{R^2-x^2}\,dx.$$

To evaluate the integral, use the substitution $$x=R\sin\theta, \quad dx=R\cos\theta\,d\theta.$$

When $x=-R$, $\theta=-\frac{\pi}{2}$, and when $x=R$, $\theta=\frac{\pi}{2}$.

Substitute into the integral: $$\int_{-R}^{R}\sqrt{R^2-x^2}\,dx = \int_{-\pi/2}^{\pi/2} \sqrt{R^2-R^2\sin^2\theta}\,(R\cos\theta)\,d\theta.$$

Simplify: $$\sqrt{R^2(1-\sin^2\theta)}=R\cos\theta.$$

Thus, $$\int_{-R}^{R}\sqrt{R^2-x^2}\,dx = \int_{-\pi/2}^{\pi/2} R^2\cos^2\theta\,d\theta.$$

Use the identity $$\cos^2\theta=\frac{1+\cos(2\theta)}{2}.$$

Then $$\int_{-\pi/2}^{\pi/2} R^2\cos^2\theta\,d\theta = \frac{R^2}{2}\int_{-\pi/2}^{\pi/2}\bigl(1+\cos(2\theta)\bigr)\,d\theta.$$

Evaluate: $$\int_{-\pi/2}^{\pi/2}1\,d\theta=\pi, \qquad \int_{-\pi/2}^{\pi/2}\cos(2\theta)\,d\theta=0.$$

So, $$\int_{-R}^{R}\sqrt{R^2-x^2}\,dx=\frac{R^2}{2}\pi.$$

Finally, multiply by $2$ to get the area of the full circle: $$A=2\cdot\frac{R^2}{2}\pi=\pi R^2.$$

$$\boxed{A=\pi R^2}$$