Question

\[\int_c y^3dx-x^3dy\] C is the circle \[x^2+y^2=4\]
\[\iint\limits_D(-3x^2-3y^2)dA\]
x=rcos\theta
y=rsin\theta
\[\int_0^2\int_0^{2\pi}-3r^3(cos^2\theta -sin^2\theta)drd\theta\]

Answer 1

so far so good I think?

Answer 2

yeah

Answer 3

too bad the sine and cosines don't cancel out as easily as in the last problem

Answer 4

no problem ... instead\[\cos^2 \theta -\sin^2 \theta=\cos 2 \theta\]

Answer 5

\[\int_0^2\int_0^{2\pi}-3r^3(\cos^2\theta -\sin^2\theta)drd\theta\]\[=\int_0^{2\pi}(\cos^2\theta -\sin^2\theta)d\theta \int_{0}^{2} -3r^3 dr =\int_0^{2\pi}(\cos 2\theta )d\theta \int_{0}^{2} -3r^3 dr\]

Answer 6

Oh sorry I've been gone all day. But yeah that makes sense. Thank you!

Answer 7

:)