curve integrals

[mathstresser]

This is a silly question, but…

\(\displaystyle \L\\\int_{c}xy^4 ds\)

C is the right half of the circle \(\displaystyle \L\\\ x^2 + y^2 = 16\)


I start of with:

\(\displaystyle \L\\\int_{3pi/2}^{pi/2} 4 dr\)

But we need to parameterize it….

How do I do that? And is that the only other thing I need to do?

 

[galactus]

\(\displaystyle \L\\x=4cos(t)\)

\(\displaystyle \L\\y=4sin(t)\)

\(\displaystyle \L\\\frac{dx}{dt}=-4sin(t)\)

\(\displaystyle \L\\\frac{dy}{dt}=cos(t)\)

\(\displaystyle \L\\ds=\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt=4\)

\(\displaystyle \L\\xy^{4}=1024cos(t)sin^{4}(t)\)

\(\displaystyle \L\\4\int_{c}(xy^{4})ds=4096\int_{0}^{\pi}(cos(t)sin^{4}(t))dt\)

Now, the fun part.

I hope I didn't err. Check it out.

 

[mathstresser]

Thanks!