bounds of integration(polar)

[akoaysigod]

Find the area of the region enclosed by one loop of r = 4sin(5t)

which is...
t - 1/10(sin10t) which needs to be integrated from ? to ?

From the picture it looks like I should be able to integrate it from 0 to pi/6. I also tried pi/3 to 2pi/3 and going from 0 to 2pi and then multiplying by 1/5. None of them yield the correct response; I am pretty sure that I integrated it correctly.

 

[DrMike]

Find the area of the region enclosed by one loop of r = 4sin(5t)

which is...
t - 1/10(sin10t) which needs to be integrated from ? to ?

From the picture it looks like I should be able to integrate it from 0 to pi/6. I also tried pi/3 to 2pi/3 and going from 0 to 2pi and then multiplying by 1/5. None of them yield the correct response; I am pretty sure that I integrated it correctly.

Well, if you want one loop, you'll need to integrate over a range of t where sin(5t) goes from 0 to 0. So I'd say the correct bounds should be 0 to pi/5, nor pi/6 (which admittedly, is very similar).

Also, how did you get t-1/10(sin10t) as the integrand?

 

[akoaysigod]

1/2 int 16sin^2(5t)
8 int sin^2(5t)
8 int sin^2(5t)
8 int 1/2 (1-cos10t)
4 int(1-cos10t)
4t - 2/5 sin(10t)

I think I factored out the 4 too soon now that I'm explaining it I think this is right and that is wrong.

 

[akoaysigod]

Also based on how it was explained in class, by one loop, it means just one loop of the 5 it draws. I am going to try what you just said and see if it works though. That was how I made my the original conclusion from going 0 to pi/6 since that looks like it encompasses just one loop.

 

[akoaysigod]

Integrating from 0 to pi/5 was correct on the corrected integrand. And I thought about it a bit more and it does encompass just one loop. Thanks.

 

[galactus]

If we set \(\displaystyle sin(5t)=0\) and solve for \(\displaystyle t=\frac{C{\pi}}{5}\)

Letting C=1, we see the limit of integration for the loop in the first quadrant.

\(\displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{5}}(4sin(5t))^{2}dt\)