Q:

Find the area bounded by the curve $y^2=4ax$ and $x=a$

Find the area bounded by the curve $y^2=4ax$ and $x=a$ 
My attempt: 
Solving equations $y^2=4ax$ and $x=a$ we get $y=\pm 2a$
And the area $=\int y. dx$ 
Is this correct?

A:

I always graph things to see what the region looks like. Using Desmos for $a=1$ this looks like the following.

We want the area between the red curve and the blue line.
There are many methods. The most straight-forward is to find the integral
$$
\int_{0}^{a} \sqrt{4ax} \,\,\mathrm{d}x
$$
which will give you the top half of the region, and then double it to get the whole region. Thus
$$
\int_{0}^{a} \sqrt{4ax} \,\,\mathrm{d}x = 2\sqrt{a} \int_{0}^{a} x^{\frac{1}{2}} \,\,\mathrm{d}x = 2\sqrt{a} \cdot\Big[ \frac{2x^{3/2}}{3} \Big]_{x=0}^{x=a} = \frac{4}{3}\cdot a^{2}.
$$
This is the top half of the region, so the area of the whole region is
$$
\frac{8}{3}\cdot a^{2}.
$$

