Finding the area of the region bounded by $x=0$, $y=0$, $2x^2=\sqrt{x^2+y^2}$, $2y^3=\sqrt{x^2+y^2}$

Question

I have 4 equations and need to find the area bounded by the corresponding curves.

I don't know how to approach it.

Well, do you know about integrals? – Théophile Sep 13 '19 at 21:44 — Théophile, Sep 13 '19 at 21:44
Find the area and square it? – Hagen von Eitzen Sep 13 '19 at 21:44 — Hagen von Eitzen, Sep 13 '19 at 21:44
sorry just finding the area under the lines. – ERJAN Sep 13 '19 at 21:48 — ERJAN, Sep 13 '19 at 21:48

score 1 · Answer 1 · answered Sep 13 '19 at 21:53

The intersection of yellow and green is when $x=y$ and $2x^3=\sqrt{2x^2}$, i.e, $x=y=2^{-1/4}\approx 0.840896$. The area below the yellow curve (i.e., bounded by its top branch (given by $y=\sqrt{4x^6-x^2}$), the $x$ axis and the line $x=2^{-1/4}$ is $$A_1=\int_{2^{-1/2}}^{2^{-1/4}}\sqrt{4x^6-x^2}\,\mathrm dx.$$ Conclude that the area of interest is $(2^{-1/4})^2-2A_1$. Now, can you compute $A_1$?

score 0 · Answer 2 · answered Sep 13 '19 at 21:58

Try using polar coordinates to simplify the problem. We have that $\sqrt{x^2+y^2} = r$ and $2x^3 = 2r^3\cos^3\theta$, so the equation $2x^3 = \sqrt{x^2+y^2}$ in polar coordinates is $r^2 = \frac{1}{2\cos^3\theta}$. Therefore, $$ A = 2\int_0^{\pi/4}\frac{1}{2}r^2d\theta = \frac{1}{2}\int_0^{\pi/4}\sec^3\theta d\theta. $$ If you don't know how to do this (rather infamous) integral, you may want to read about various techniques here. Hint: try integration by parts.

Finding the area of the region bounded by $x=0$, $y=0$, $2x^2=\sqrt{x^2+y^2}$, $2y^3=\sqrt{x^2+y^2}$

2 Answers2