1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ desmos.com/calculator/ztvwh65pyj. Are you looking for the region thats looks like a triangle bounded above by y=6? $\endgroup$ – Ty. Jun 2 '20 at 17:32
  • $\begingroup$ @Ty. Yes, I think so. The triangle upper-bounded by $y=6$. $\endgroup$ – Daniel Jun 2 '20 at 17:33
1
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ – Daniel Jun 2 '20 at 18:16
  • $\begingroup$ 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? $\endgroup$ – José Carlos Santos Jun 2 '20 at 18:19
  • $\begingroup$ 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. $\endgroup$ – Daniel Jun 2 '20 at 18:24
  • $\begingroup$ No. If it's only that upper triangle that you are interested in, how can those integrals begin at $0$? $\endgroup$ – José Carlos Santos Jun 2 '20 at 18:26
  • $\begingroup$ 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 $\endgroup$ – Daniel Jun 2 '20 at 18:29
2
$\begingroup$

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

enter image description here

$\endgroup$
2
  • $\begingroup$ Can you explain, please, $\sqrt{x}$, as lower boundary in inner integrals? $\endgroup$ – zkutch Jun 3 '20 at 1:05
  • $\begingroup$ @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 $\sqrt{4x}$. $\endgroup$ – user9464 Jun 3 '20 at 21:47
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.