Question

Using double integral fund the area of region enclosed by √x+√y=√a and x+y=a

Answer 1

Area bounded by the two curves = $\displaystyle\iint_{R}dydx$.

The limits of the integral are obtained from the given equations.

$\sqrt{x} + \sqrt{y} = \sqrt{a} \implies\sqrt{y} = \sqrt{a}-\sqrt{x} \implies y = (\sqrt{a}-\sqrt{x})^2\\ x+y=a \implies y = a-x$

The limits for x are obtained from the two equations of the curve.

$\begin{aligned} (\sqrt{a} - \sqrt{x})^{2} &= a - x\\ a - 2\sqrt{ax} + x &= a - x\\ 2x &= 2\sqrt{ax}\\ 4x^{2} &= 4ax \quad\text{(Squaring both sides)}\\ 4x^{2} - 4ax &= 0\\ 4x(x-a) &= 0 \end{aligned}$

which gives, $x = 0~ \&~ x=a$.

Hence, y varies from $y = (\sqrt{a}-\sqrt{x})^2 \text{ to } y = a-x$ and x varies from $x = 0 \text{ to } x = a.$

Therefore,

$\begin{aligned} \text{Area bounded by the curves } &= \displaystyle\int_{0}^{a}\int_{(\sqrt{a}-\sqrt{x})^2}^{a-x}dydx\\ &= \displaystyle\int_{0}^{a} y\Bigg|_{(\sqrt{a}-\sqrt{x})^2}^{a-x} dx\\ &= \displaystyle\int_{0}^{a} ((a-x)-(\sqrt{a}-\sqrt{x})^2)dx\\ &= \displaystyle\int_{0}^{a} (a - x - a - x +2\sqrt{ax})dx\\ &= \displaystyle\int_{0}^{a} ( - 2x + 2\sqrt{ax})dx\\ &= 2\displaystyle\int_{0}^{a} ( \sqrt{ax} - x)dx\\ &= 2 \Big(\sqrt{a} ~\frac{x^{\frac{3}{2}}}{(\frac{3}{2})} - \frac{x^2}{2}\Big)\Bigg|_0^a\\ &= 2 \Big(\frac{2}{3}\sqrt{a}\cdot a^{\frac{3}{2}} - \frac{a^2}{2}\Big)\\ &= 2 \Big(\frac{2a^2}{3} - \frac{a^2}{2}\Big)\\ &= \frac{a^2}{3} \end{aligned}$