<s>[INST] Calculation of area in 2 definite integrals given function $y=x^2$ Here is a graph for $y=x^2$

Given that the area in blue is equal to the area in pink, find a in terms of b and solve for a.
My attempt:
From the graph I can see that :$a^2=b$ and $a=\sqrt b$
Since area blue is equal to area in pink:
$$\int^a_1 x^2 dx=\int^b_1 y^\frac{1}{2}dy$$
$$\left[\frac{1}{3}x^3\right]^a_1=\left[\frac{2}{3}y^\frac{3}{2}\right]^b_1$$
$$\frac{1}{3}a^3-\frac{1}{3}=\frac{2}{3}b^\frac{3}{2}-\frac{2}{3}$$
$$a^3-1=2b^\frac{3}{2}-2$$
$$a^3-1=2a^3-2$$
$$a^3=1$$
$$a=1$$
The answer to $a=1$ makes sense because when $a=1$, the area becomes $0$    $units^2$
But my teacher said the answer is $a=1+\sqrt 3$
Is there any idea what has gone wrong in my calculations?
Also please tell me how to get to the answer as well?
 [/INST] Note$$\bigg(\int_{1}^{1+\sqrt{3}} x^2 dx\bigg) -\bigg(\int_{1}^{(1+\sqrt{3})^2} \sqrt{y} dy \bigg) = -(3+2\sqrt{3}) \neq 0$$
So your teacher is wrong in this case, and you are right.
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