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

Question

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?

Answer 1

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 $$ \frac{8}{3}\cdot a^{2}. $$

Answer 2

If you swich both axsis you can figure out you can do like this:

$$Area = 4a^2-2\int _0^{2a}{y^2\over 4a}dy = 4a^2-2{y^3\over 12a}\Big|_0^{2a}$$ $$= 4a^2- 2{8a^3\over 12a}= 4a^2 -4a^2/3 =8a^2/3$$

Answer 3

Solving $y^2=4ax$ and $y=6a$ we get $x=9a$ so our integral will be $2\int_{0}^{9a}\sqrt{4ax}dx$