1. The problem statement, all variables and given/known data find the region enclosed by astroid x= acosϑ ^{3} y=asinϑ ^{3} the astroid looks like this picture in my book, with the shaded region stretching for x (-a,a) and for y(-a,a) 2. Relevant equations i think the formula is ..... from (-a to a) ∫ y * x' dϑ 3. The attempt at a solution a = ∫ acosϑ ^{3} * 3asinϑ ^{2} cosϑ a = 3a^{2}∫ cosϑ ^{4} * sinϑ ^{2} a = 3a^{2}∫ cosϑ ^{4} * (1-cosϑ ^{2} ) a = 3a^{2}∫ cosϑ ^{4} - cosϑ^{6} is this one the right track, if so I can evaluate the rest of that integral myself
Your formula would be: [tex]A=\int_{-a}^a y dx = \int_{\pi}^0 y \frac {dx} {d\theta} d\theta[/tex] Please note that this will be only the area above the x-axis. To get the total area, you will need to double that. Your derivative of x should be: [tex]\frac {dx} {d\theta}=\frac {d} {d\theta} (a\cos^3\theta)=3a\cos^2\theta\cdot -\sin\theta[/tex] Substituting gives: [tex]A=\int_{\pi}^0 (a\sin^3\theta) \cdot (3a\cos^2\theta\cdot -\sin\theta) d\theta =3a^2\int_0^{\pi} \sin^4\theta \cos^2\theta d\theta [/tex] I think you can take it from there. So you were on the right track, but made a mistake in substitution. [edit]Also, you should take note of the integral boundaries. They can not just be left out.[/edit]
quick question, how did you decide the bound from pi to 0 and then you flipped them, it has something to do with the negative sign right ?
In the first equation we shift from x to theta. A value of x=-a corresponds to a value theta=pi, and x=+a corresponds to theta=0. So the integral is from pi to 0. In the third equation, there is a minus sign in the equation. Flipping the boundaries means flipping the sign, or in this case removing the minus sign.
okay i was scratching my head about this one but i think you get it by just substituting x for a and solving for theta right? -a = acos^{3}ϑ -1 = cos^{3}ϑ ϑ = pi a = acos^{3}ϑ 1 = cos^{3}ϑ ϑ = 0 did i arrive at that correctly?
Yes. That is entirely correct! Actually, what I did is look at the graph you posted. With theta being the angle with the positive x-axis, a value of pi corresponds to x=-a. Substituting pi in the expression for x shows that this is correct.