# Area of astroid given in parametric form

Find the area of the region bounded by the astroid, in parametric form, $$(x;y)=(2cos^3 t; 2sin^3 t)$$

Well, I used the formula of area given in parametric curves $$\int_a^b y(t).x'(t) \,dt$$. So, as it's an astroid, I know that I can find the area between $$[0; \pi ]$$ and multiply it by 2. Then, after differentiating and replacing, I get $$2 \int_0^\pi 2sin^3(t).(-6cos^2(t)sin(t)) \,dt$$

which can be written as $$-24 \int_0^\pi sin^4(t).cos^2(t) \,dt$$

The thing is that I integrated using trigonometric identities, but it was really tedious, and, to top it, it was wrong, because when I differentiated the result I didn't get the first function and the area was negative.

• – Dr. Sonnhard Graubner Jul 6 at 14:45
• You might have an easier time of it if you use $\frac12(x\,dy-y\,dx)$ as the integrand. – amd Jul 6 at 18:02
• @Dr.SonnhardGraubner I’m not sure that question helps much. The OP is already trying to use the method given in the accepted answer; the issue appears to be in the details of the actual calculation. – amd Jul 6 at 18:05
• The sign of your answer came out wrong because you left out a negative sign in the first place. As to why the value is otherwise incorrect, you’ll have to show your work for anyone to be able to point out your error. – amd Jul 6 at 18:17

We can dispense with the reason that you ended up with a negative area easily: you left a negative sign out in the first place. Referring to this answer to an almost-identical question, the area of the astroid within the first quadrant is $$-\int_0^{\pi/2} x'(t) y(t) \,dt$$. The negative sign appears because $$x$$ decreases as $$t$$ increases from $$0$$ to $$\frac\pi2$$.
As for the rest, it’s impossible to say why you ended up with the wrong value without seeing your work. However, I think you could’ve made it a bit easier on yourself by using a different area element. The symmetries of $$\sin$$ and $$\cos$$ to me suggest using a more balanced volume element, namely $$\frac12(x\,dy-y\,dx)$$, or $$\frac12 \begin{vmatrix}x(t)&y(t)\\x'(t)&y'(t)\end{vmatrix} dt = \frac12\left(x(t)y'(t)-y(t)x'(t)\right) dt.$$ You can visualize this as the area of the infinitesimal triangle with sides defined by the vectors $$(x,y)$$ and $$(x+dx,y+dy)$$, i.e., an approximation to the area swept out by the radius vector $$\mathbf r(t)=\left(x(t),y(t)\right)$$ between $$t$$ and $$t+dt$$. For the astroid, the integrand is therefore \begin{align}\frac12 \begin{vmatrix}2\cos^3t & 2\sin^3t \\ -6\cos^2t\sin t & 6\cos t\sin^2t\end{vmatrix} &= 6\left(\cos^4t\sin^2t+\cos^2t\sin^4t\right) \\ &= 6\cos^2t\sin^2t \\ &= \frac32\sin^2{2t}, \end{align} which can be simplified further to $$\frac34(1-\cos{4t})$$ using the identity $$\cos{2\theta} = 1-2\sin^2\theta$$.