Calculating Area of a Polar Graph with One Loop | Take Home Test Extra Credit

[OrbsAndSuch]

This is an extra credit problem for a take home test, so i will understand if no one feels comfortable helping me out, but any advice is greatly appreciated

Homework Statement

Compute the area enclosed by one loop of the graph given by r = sqrt(sin(3{theta}))

Homework Equations

see above

The Attempt at a Solution

The graph makes 3 loops, so i tried finding the area from (0, 2{pi}) but all come up with is 0


Thanks again!

 

[Mark44]

You get one loop as theta ranges from 0 to pi/3.

 

[OrbsAndSuch]

Thank you Mark. Is finding the finding the area of the entire graph, then dividing that answer by the number of loops a viable method for this type of problem?

 

[Mark44]

You could do it that way, I suppose, but it makes more sense to me to get the area within one loop and multiply it by the number of loops. Keep in mind that the integrand is undefined for theta in [pi/3, 2pi/3], because of the square root.

 

[OrbsAndSuch]

The formula for the area [ A = (1/2)*r² ] effectively eliminates the square root, so i don't quite understand how its undefined from [pi/3, 2pi/3] because of the square root.

However after proving to myself that r = 0 at 0 and pi/3, I have successfully found the area of one loop. Thanks again for all the help Mark

 

[Mark44]

I was just looking at r in your first post, and wasn't thinking about the integral.