Finding the Area Enclosed by a Polar Curve

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Can someone please help me on this question. I tried to solve it by integrating 0.5*(1-3sin(θ)^2 from -Pi/2 to 0 but I didnt get the answer.
 

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amninder15 said:
Can someone please help me on this question. I tried to solve it by integrating 0.5*(1-3sin(θ)^2 from -Pi/2 to 0 but I didnt get the answer.

Why did you choose ##-\frac \pi 2## to ##0##? Just guessing? Have you drawn a sketch? Do you know what ##\theta## give negative ##r## values for the inner loop?
 
Yea I made some silly mistake I think It goes from sin^-1(1/3) to Pi-sin^-1(1/3). But that looks ugly. Am I on right path.
 

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Your graph is correct for ##\theta## varies between 0 and 2pi. Now, generally how do you find the area of a polar region?
 
amninder15 said:
Yea I made some silly mistake I think It goes from sin^-1(1/3) to Pi-sin^-1(1/3). But that looks ugly. Am I on right path.

Yes. Those are the correct limits. Ugly or not, just plow ahead and you will get one of the answers listed.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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