
#1
Apr1511, 05:20 PM

P: 187

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} * (1cosϑ ^{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 



#2
Apr1611, 10:43 AM

HW Helper
P: 6,189

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 xaxis. 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] 



#3
Apr1811, 10:12 AM

P: 187

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
? 



#4
Apr1811, 10:19 AM

HW Helper
P: 6,189

find the area of region enclosed by astroidA 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. 



#5
Apr1811, 10:27 AM

P: 187

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? 


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