Find the area bounded by $x= at^{2}$ and $y = 2at$ from $t=1$ to $t=2$

Question

Find the area bounded by $x= at^{2}$ and $y = 2at$ from $t=1$ to $t=2$

I tried to solve this by integrating $\int_{1}^2 y \frac{dx}{dt} dt$

$\int_{1}^2 (4a^{2}t^{2}) dt$

$= (28/3)a^{2}$

What is wrong with my attempt$?$

Answer is given $(56/3)a^{2}$

Answer 1

Hint: You can also use an explicit form of your function: $$t=\sqrt{\frac{x}{a}}=t$$ so $$y=2a\sqrt{\frac{x}{a}}$$

Answer 2

Your answer is just fine, because $$\int_{1}^{2} 4a^2 ~dt = \int_{a}^{4a} 2a\sqrt{\frac{x}{a}}~ dx = 28a^2/3$$ is the area between the parametric curve and the $x(t)$-axis.

Answer 3

The curve is a parabola y^2 = 4aX area will be twice the area under curve y=sqrt.(4ax) from X= a to X = 4a and x axis answer is 56 a^2 / 3