Calculate area using double integral I'm trying to calculate the area defined by the following curves: $y=x^2, 4x=y^2, y=6$ using double integrals.
I'm wondering whether my solution is correct:
Area = $\int^{6}_{0}\int^{\sqrt{4x}}_{0}1dydx - \int^{6}_{0}\int^{x^2}_{0}1dydx$
Is it correct?
Thanks!
 A: Drawing a picture is helpful to set up the double integral properly.
It is not difficult the find out the coordinates of the points $A, B, C, D$ by solving relevant equations. Let's denote $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ and $D=(d_1,d_2)$. Then one way to set up the double integral is
$$
\int_{b_1}^{c_1}\left(\int_{\sqrt{4x}}^{x^2}1\; dy\right)\;dx+
\int_{c_1}^{d_1}\left(\int_{\sqrt{4x}}^{6}1\; dy\right)\;dx\;.
$$

A: No, that is not correct.
First, you should compute the intersection points of the curves $y=x^2$ and $4x=y^2$, which are $(0,0)$ and $\left(\sqrt[3]4,2\sqrt[3]2\right)$. When $y\in\left[0,2\sqrt[3]2\right)$, the curve $y=x^2$ is to the right of the curve $4x=y^2$; after that, it is located to the left.
So, you should compute$$\int_0^{2\sqrt[3]2}\int_{y^2/4}^{\sqrt y}1\,\mathrm dx\,\mathrm dy+\int_{2\sqrt[3]2}^6\int^{y^2/4}_{\sqrt y}1\,\mathrm dx\,\mathrm dy.$$
A: Second possible answer, additional to José Carlos Santos's one, is
$$\int_{ \sqrt[\leftroot{-2}\uproot{2}3]{4}}^{\sqrt{3}} \int_{\sqrt{4x}}^{x^{2}} dxdy + \int_{\sqrt{3}}^{9} \int_{\sqrt{4x}}^{6}dxdy $$
For your variant it should be
$$\int^{6}_{0}\int^{\frac{y^2}{4}}_{0}1dydx - \int^{2\sqrt[\leftroot{-2}\uproot{2}3]{2}}_{0}\int^{\frac{y^2}{4}}_{0}1dydx - \int_{2\sqrt[\leftroot{-2}\uproot{2}3]{2}}^{6} \int_{0}^{\sqrt{y}}dydx $$
But there is 4-"angle" figure in the left of 3-"angle", which is also bounded by 3 curves:
$$ \int_{-\sqrt{3}}^{0} \int_{x^2}^{6} dxdy  + \int_{0}^{\sqrt[\leftroot{-2}\uproot{2}3]{4}} \int_{\sqrt{4x}}^{6} dxdy + \int_{\sqrt[\leftroot{-2}\uproot{2}3]{4}}^{\sqrt{3}} \int_{x^2}^{6} dxdy $$
