# Area bounded by hypocycloid

1. May 5, 2006

### Reshma

Equation of a hypocycloid is:
$$x^{3/2} + y^{3/2} = a^{3/2}$$.
Find the area of the figure bounded by this hypocycloid.

My work:
I can use the plane polar coordinates here taking $x = a\cos t$ & $y = a\sin t$ with $t = [0, 2\pi]$. But I don't know how to obtain the surface integral for evaluating the area ( :yuck: ) . Someone help me.

2. May 5, 2006

### siddharth

First of all, you've not used plane polar coordinates. When you say $x = a\cos t$ & $y = a\sin t$, you have just given a parametric representation of any point (x,y) lying on the curve.

When you integrate to find the area, your domain of integration is going to include points lying inside the curve also. So, you need to use the polar coordinates $x= r \cos \theta$ and $y=r \sin \theta$.

$$\int \int dx dy$$
Convert this integral to polar coordinates. Can you take it from here?

Last edited: May 5, 2006
3. May 6, 2006

### Reshma

Well in any case $\int \int dx dy$ the cannot be applied here since it completely ignores the equation of the given curve. The parametric representation of a hypocycloid will be:
$$x = a\cos^3 t , y = a\sin^3 t$$
$$dx = -3a\cos^2 t\sin tdt, dy = 3a\sin^2 t\cos t dt$$
So the formula representing the area of a curvilinear trapezoid bounded by a curve represented parametrically is:
$$Q = \int_0^{2\pi} ydx = \int_0^{2\pi} a\sin^3 t(-3a\cos^2 t\sin t)dt =-3a^2\int_0^{2\pi} \sin^4 t\cos^2 tdt$$

Someone just help me crack this integral, I am not very far from the answer i.e. ${3\over 8}\pi a^2$.

4. May 6, 2006

### HallsofIvy

Staff Emeritus
Looks to me like a standard "even powers of trig functions". Use the identities: sin2x= (1/2)(1- cos(2x)) and cos2= (1/2)(1+ cos(2x)) to reduce the powers.

5. May 6, 2006

### 0rthodontist

These do not agree (plug in with e.g. t = 2, a = 1 to verify). You want your x to be such that x^(3/2) = a^(3/2) * cos^2 t and your y to satisfy y^(3/2) = a^(3/2) * sin^2 t.

Or is your original equation correct? It would only be defined for values in the first quadrant.

Last edited: May 7, 2006