Area — Question 1

Question 1

Find the exact area of the region enclosed by the curves $$y = x\sqrt{4-x^2} \quad\text{and}\quad y = x^2.$$

Question 1 - Solution

First find the points of intersection: $$x\sqrt{4-x^2} = x^2.$$

Factor: $$x\bigl(\sqrt{4-x^2}-x\bigr)=0.$$

Thus, $$x=0 \quad\text{or}\quad \sqrt{4-x^2}=x.$$

Square the second equation: $$4-x^2=x^2 \quad\Rightarrow\quad x^2=2 \quad\Rightarrow\quad x=\sqrt{2}.$$

On the interval $[0,\sqrt{2}]$, we have $$x\sqrt{4-x^2} \ge x^2.$$

Therefore, the area is $$A=\int_{0}^{\sqrt{2}} \left(x\sqrt{4-x^2}-x^2\right)\,dx.$$

Split the integral: $$A=\int_{0}^{\sqrt{2}} x\sqrt{4-x^2}\,dx -\int_{0}^{\sqrt{2}} x^2\,dx.$$

For the first integral, use substitution. Let $$u=4-x^2, \qquad du=-2x\,dx.$$

Then $$\int x\sqrt{4-x^2}\,dx = -\frac12\int u^{1/2}\,du = -\frac13 u^{3/2} = -\frac13(4-x^2)^{3/2}.$$

Evaluate from $0$ to $\sqrt{2}$: $$\left[-\frac13(4-x^2)^{3/2}\right]_{0}^{\sqrt{2}} = -\frac13(2\sqrt{2})+\frac13(8) = \frac{8-2\sqrt{2}}{3}.$$

For the second integral: $$\int_{0}^{\sqrt{2}} x^2\,dx = \left[\frac{x^3}{3}\right]_{0}^{\sqrt{2}} = \frac{2\sqrt{2}}{3}.$$

Subtract: $$A=\frac{8-2\sqrt{2}}{3}-\frac{2\sqrt{2}}{3} =\frac{8-4\sqrt{2}}{3}.$$

$$\boxed{ \frac{8-4\sqrt{2}}{3} }$$