The area bounded by the parabola $y²=4ax$ and straight line $x+y=3a$ is....?
I just drew the graph and got two intersecting area , one with $+ve$ y-axis and above parabola and another with $+ve$ x-axis and below parabola.. then find the area using double integral and got the ans $11a²/3$, but answer given is $10a²/3$...
I disagree, the intercepts are $(a, 2a)$ and $(9a,-6a)$ so the integral is (assuming $a>0$):
$$\begin{align}\int_{-6a}^{2a}\int_{y^2/4a}^{3a-y}1\operatorname d x\operatorname d y ~=~& \int_{-6a}^{2a}3a-y-\frac{y^2}{4a} \operatorname d y \\[1ex]~=~& \Big(3ay-\frac{y^2}2-\frac{y^3}{12a}\Big)\Big\rvert_{y=-6a}^{y=2a}\\[1ex]~=~&\frac{64a^2}{3}\end{align}$$
Wait: the area bounded by the curves and above the x-axis is (assuming $a>0$):
$$\begin{align}\int_{0}^{2a}\int_{y^2/4a}^{3a-y}1\operatorname d x\operatorname d y ~=~&\frac{10a^2}{3}\end{align}$$