Area enclosed by r = 1 - 3sin²θ

[amninder15]

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.

 

[LCKurtz]

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?

 

[amninder15]

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.

 

[DryRun]

Your graph is correct for $\theta$ varies between 0 and 2pi. Now, generally how do you find the area of a polar region?

 

[LCKurtz]

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.