Question

Find by double integration the area of the region in 𝑥𝑦 plane bounded by the curves 𝑦 = 𝑥

2 and

𝑦 = 4𝑥 − 𝑥

2

.

Answer 1

$$x^2=4-x^2$$

$$x_1=-\sqrt{2}, x_2=\sqrt{2}$$

$$A=\displaystyle\int_{-\sqrt{2}}^{\sqrt{2}}\displaystyle\int_{x^2}^{4-x^2}dydx$$

$$=\displaystyle\int_{-\sqrt{2}}^{\sqrt{2}}[y]\begin{matrix} 4-x^2 \\ x^2 \end{matrix}dx$$

$$=\displaystyle\int_{-\sqrt{2}}^{\sqrt{2}}(4-2x^2)dx$$

$$=[4x-\dfrac{2x^3}{3}]\begin{matrix} \sqrt{2} \\ - \sqrt{2} \end{matrix}$$

$$=4 \sqrt{2}-\dfrac{4\sqrt{2}}{3}+4\sqrt{2}-\dfrac{4\sqrt{2}}{3}$$

$$=\dfrac{16\sqrt{2}}{3} ({units}^2)$$