# Line integral over half circle.. I'm stuck!

• Nov 2nd 2012, 04:25 PM
SparrowTail
Line integral over half circle.. I'm stuck!
Hello ppl!

So I'm frustrated with this problem:Evaluate the line integral , where C is the right half of the circle x2+y​2=36

So far, I've taken x=6Cos(t) and y = 6Sin(t) to parametrize a circle with radius of 6, then taken their modulus (to get rid of ds and get dt) and got 6 because rest of it cancels out..
After that I replaced the xy2 with the trig. values to get 5*6*Cos(t)Sin2(t). After that when I integrate it with respect to t, 3Pi/2 < t < Pi/2 to go over only the right side of the circle, I'm getting 20, which is wrong because the browser is rejecting it as the wrong answer.. :/
I think I'm doing something really silly. Anyone wanna help?

Thanks!
-Sparrow
• Nov 2nd 2012, 04:37 PM
Prove It
Re: Line integral over half circle.. I'm stuck!
Quote:

Originally Posted by SparrowTail
Hello ppl!

So I'm frustrated with this problem:Evaluate the line integral , where C is the right half of the circle x2+y​2=36

So far, I've taken x=6Cos(t) and y = 6Sin(t) to parametrize a circle with radius of 6, then taken their modulus (to get rid of ds and get dt) and got 6 because rest of it cancels out..
After that I replaced the xy2 with the trig. values to get 5*6*Cos(t)Sin2(t). After that when I integrate it with respect to t, 3Pi/2 < t < Pi/2 to go over only the right side of the circle, I'm getting 20, which is wrong because the browser is rejecting it as the wrong answer.. :/
I think I'm doing something really silly. Anyone wanna help?

Thanks!
-Sparrow

The region \displaystyle \begin{align*} \frac{\pi}{2} < t < \frac{3\pi}{2} \end{align*} is actually the LEFT side of the circle. You should use \displaystyle \begin{align*} -\frac{\pi}{2} < t < \frac{\pi}{2} \end{align*}
• Nov 2nd 2012, 04:42 PM
SparrowTail
Re: Line integral over half circle.. I'm stuck!
Thanks for the help but that just gives me 20 again! :P
• Nov 2nd 2012, 04:51 PM
Prove It
Re: Line integral over half circle.. I'm stuck!
I think you'll find \displaystyle \begin{align*} 5\,x\,y^2 = 5 \left( 6\cos{t} \right) \left( 6\sin{t} \right)^2 = 30\cos{t} \cdot 36\sin^2{t} = 720\cos{t}\sin^2{t} \end{align*}
• Nov 2nd 2012, 05:02 PM
SparrowTail
Re: Line integral over half circle.. I'm stuck!
Gosh! totally screwed that one didn't I? Thankfully I have unlimited tries for this one...
Okay So I tried it with 720 Cos(t)Sin^2(t), but still the wrong answer..
Also, (at the risk of sounding stupid), I did it with 1080Cos(t)Sin^2(t) because 30*36 = 1080..
But still no luck.. Thank you so much for your help though..
• Nov 2nd 2012, 05:06 PM
Prove It
Re: Line integral over half circle.. I'm stuck!
Quote:

Originally Posted by SparrowTail
Gosh! totally screwed that one didn't I? Thankfully I have unlimited tries for this one...
Okay So I tried it with 720 Cos(t)Sin^2(t), but still the wrong answer..
Also, (at the risk of sounding stupid), I did it with 1080Cos(t)Sin^2(t) because 30*36 = 1080..
But still no luck.. Thank you so much for your help though..

Yes I really should learn to multiply.

Did you remember to times this by 6 from your ds?
• Nov 2nd 2012, 05:09 PM
SparrowTail
Re: Line integral over half circle.. I'm stuck!
DANG! That's what was wrong! Lot's of n00b mistakes in here on my part... Gosh thank you so much for your help, kind person!
Saved me a night's sleep you did! :)