# 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!

• desmos.com/calculator/ztvwh65pyj. Are you looking for the region thats looks like a triangle bounded above by y=6? – Ty. Jun 2 '20 at 17:32
• @Ty. Yes, I think so. The triangle upper-bounded by $y=6$. – Daniel Jun 2 '20 at 17:33

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$$\int_0^{2\sqrt[3]2}\int_{y^2/4}^{\sqrt y}1\,\mathrm dx\,\mathrm dy+\int_{2\sqrt[3]2}^6\int^{y^2/4}_{\sqrt y}1\,\mathrm dx\,\mathrm dy.$$

• Are you sure it's correct? I'm not sure about the first integral. As far as I understand, you are calculating the area of the following part: i.imgur.com/CqHJVPM.png but it should be just upper "triangle". Shoudn't it just be the second integral? – Daniel Jun 2 '20 at 18:16
• If you are sure that it's just the upper triangle, then you're right, of course. Do you want me to edit my answer? – José Carlos Santos Jun 2 '20 at 18:19
• Oh I see now. No, that's not necessary. Thanks. Just a quick question, considering only the triangle, is my solution mentioned in the post correct? Just want to know if I understand it correctly. – Daniel Jun 2 '20 at 18:24
• No. If it's only that upper triangle that you are interested in, how can those integrals begin at $0$? – José Carlos Santos Jun 2 '20 at 18:26
• My way of thinking about it was that I take this area: i.imgur.com/oPSqKjt.png and subtract this area: i.imgur.com/57IK5l4.png – Daniel Jun 2 '20 at 18:29

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 $$\int_{b_1}^{c_1}\left(\int_{\sqrt{4x}}^{x^2}1\; dy\right)\;dx+ \int_{c_1}^{d_1}\left(\int_{\sqrt{4x}}^{6}1\; dy\right)\;dx\;.$$

• Can you explain, please, $\sqrt{x}$, as lower boundary in inner integrals? – zkutch Jun 3 '20 at 1:05
• @zkutch: thank you for identifying a typo! The lower boundary should come from the curve $4x=y^2$ ($y>0$) and thus it should be $\sqrt{4x}$. – user9464 Jun 3 '20 at 21:47

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: $$\int_{-\sqrt{3}}^{0} \int_{x^2}^{6} dxdy + \int_{0}^{\sqrt[\leftroot{-2}\uproot{2}3]{4}} \int_{\sqrt{4x}}^{6} dxdy + \int_{\sqrt[\leftroot{-2}\uproot{2}3]{4}}^{\sqrt{3}} \int_{x^2}^{6} dxdy$$