Question

Prove that the area of a circle of radius $r\text{is}\pi r^2$, as follows:

(a) Write down a definite integral that represents the area of the circle of radius $r$centered at the origin. (Hint: The equation of such a circle is $x^2+y^2=r^2$.)

(b) Use trigonometric substitution to solve the definite integral you found in part (a). (Hint: Change the limits of integration to match your substitution .)

Short answer

Part (a) The integral is $2{\int }_{-r}^r\sqrt{r^2-x^2}dx$.

Part (b) The value is$\pi r^2\text{.}$

Step-by-step solution

Part (a) Step 1. Given information.

The equation of the circle is$x^2+y^2=r^2$.

Part (a) Step 2. Explanation.

The definite integral will be,

$$\begin{gathered} x^2+y^2=r^2 \\ {y}^2=r^2-x^2 \\ y=\sqrt{r^2-x^2} \\ 2{\int }_{-r}^r\sqrt{r^2-x^2}dx \end{gathered}$$

Part (b) Step 1. Explanation.

On solving,

$$\begin{gathered} 2{\int }_{-r}^r\sqrt{r^2-x^2}dx \\ =2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{r^2-\left(r^2{\sin }^{2}u\right)}\left(r\cos u\right)du \\ =2r^2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\cos }^{2}udu \\ =2r^2\left(\frac{1}{2}\right){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(1+\cos 2u)du \\ =r^2{\left[u+\frac{1}{2}\sin \left(2u\right)\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \\ ={\mathrm{\pi r}}^2 \end{gathered}$$