# Thread: Find the area bounded by...

1. ## Find the area bounded by...

Find the area bounded by,

$\displaystyle (x^{2}+y^{2})^{3} = 4a^{2}x^{2}y^{2}$

I'm a little confused by the question itself. What are they talking about? Is it a surface? It it simply something in the x-y plane? How do I get this one started?

Thanks again!

2. Originally Posted by jegues
Find the area bounded by,

$\displaystyle (x^{2}+y^{2})^{3} = 4a^{2}x^{2}y^{2}$

I'm a little confused by the question itself. What are they talking about? Is it a surface? It it simply something in the x-y plane? How do I get this one started?

Thanks again!
Use the polar coordinates

$\displaystyle \displaystyle{(x^2+y^2)^3=4a^2x^2y^2}$

$\displaystyle \displaystyle{x=\rho\cos\varphi,~y=\rho\sin\varphi }$

$\displaystyle \displaystyle{\rho^6=4a^2\rho^4\cos^2\varphi\sin^2 \varphi}$

$\displaystyle \displaystyle{\rho^2=4a^2\cos^2\varphi\sin^2\varph i}$

$\displaystyle \displaystyle{\rho^2=a^2\sin^22\varphi}$

$\displaystyle \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet a\rho^2\,d\varphi=\frac{a^2}{2}\int\limits_0^{2\pi }\sin^22\varphi\,d\varphi=\ldots=\frac{a^2\pi}{2}}$

P.S. $\displaystyle \rho=a\sin2\varphi$ is the equation of the polar rose with four petals (see wikipedia).

3. Originally Posted by DeMath
Use the polar coordinates

$\displaystyle \displaystyle{(x^2+y^2)^3=4a^2x^2y^2}$

$\displaystyle \displaystyle{x=\rho\cos\varphi,~y=\rho\sin\varphi }$

$\displaystyle \displaystyle{\rho^6=4a^2\rho^4\cos^2\varphi\sin^2 \varphi}$

$\displaystyle \displaystyle{\rho^2=4a^2\cos^2\varphi\sin^2\varph i}$

$\displaystyle \displaystyle{\rho^2=a^2\sin^22\varphi}$

$\displaystyle \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet a\rho^2\,d\varphi=\frac{a^2}{2}\int\limits_0^{2\pi }\sin^22\varphi\,d\varphi=\ldots=\frac{a^2\pi}{2}}$

P.S. $\displaystyle \rho=a\sin2\varphi$ is the equation of the polar rose with four petals (see wikipedia).
I have one quick question regarding this line,

$\displaystyle \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet a\rho^2\,d\varphi$

Where does this come from? Is this a known formula for area or something?

4. Originally Posted by jegues
I have one quick question regarding this line,

$\displaystyle \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet a\rho^2\,d\varphi$

Where does this come from? Is this a known formula for area or something?
Yes

See Integral calculus (Area) Polar coordinate system - Wikipedia, the free encyclopedia