Evaluate the line integral where $C$ is the given curve:
$\int_C xy^4 ds $, $C$ it the right half of the circle $x^2+y^2=16$
I was following a similar example in my book and parameterized $t$ by making $x = \cos t$ and $y = \sin t$ and using the line integral equation, but I got a small fraction while the answer is a large number. I looked at the example twice and checked my problem twice, so I'm not sure what I'm missing. If anyone could help it would be greatly appreciated..
$\endgroup$ 63 Answers
$\begingroup$You forgot to include the radius. The substitution $x= \cos t$, $y= \sin t$ parametrizes a circle of radius $1$.
But $C$ is a subset of a circle of radius $4$.
$$r(t)=(4\cos t,\,4\sin t)\;,\;\;-\frac\pi2\le t\le\frac\pi2\implies \text{our integral is}$$
$$\int_{-\frac\pi2}^{\frac\pi2}4\cos t(4\sin t)^4\left\|r'(t)\right\|dt=1,024\int_{-\frac\pi2}^{\frac\pi2}\cos t\,\sin^4t\cdot4\,dt=\left.\frac{4,096}5\sin^5t\right|_{-\frac\pi2}^{\frac\pi2}=\frac{8,192}5$$
$\endgroup$ 3 $\begingroup$The correct parameterization is $x=4\cos(t), y=4\sin(t)$.
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