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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.

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  • $\begingroup$ See here math.stackexchange.com/questions/66027/… $\endgroup$ – Dr. Sonnhard Graubner Jul 6 at 14:45
  • $\begingroup$ You might have an easier time of it if you use $\frac12(x\,dy-y\,dx)$ as the integrand. $\endgroup$ – amd Jul 6 at 18:02
  • $\begingroup$ @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. $\endgroup$ – amd Jul 6 at 18:05
  • $\begingroup$ 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. $\endgroup$ – amd Jul 6 at 18:17
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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 $\cos{2\theta} = 1-2\sin^2\theta$.

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