Question

Evaluate the line integral (x^2 - y)ds over the circle C: x^2 + y^2 = 4 in the first quadrant from point (0,2) to (sqrt2, sqrt2).

Answer 1

parametrize the circle using \(\vec{r}(t)=(2\sin t,2\cos t)\) so that \(\vec{r}(0)=(0,2)\) and \(\vec{r}\left(\frac\pi4\right)=(\sqrt2,\sqrt2)\). Observe that we can then rewrite our integral:$$\int_C (x^2-y)\,ds=\int_0^\frac\pi4 (x^2-y)|\vec{r}'(t)|\,dt$$

Answer 2

Since \(\vec{r}'(t)\) is easily found to be \((2\cos t,-2\sin t)\) we have \(\|\vec{r}'(t)\|=2\sqrt{\cos^2 t+\sin^2 t}=2\), and also substitute \(x=-2\sin t,y=2\cos t\) to yield:$$\int_C(x^2-y)\,ds=2\int_0^\frac\pi4(4\sin^2 t-2\cos t)\,dt$$

Answer 3

oh ok

Answer 4

Observe that we can reduce our integral into a 'nicer' form using the following trigonometric identity (if you want a derivation just ask):$$\sin^2 t=\frac12(1-\cos2t)$$thus our integral becomes:$$\begin{align*}\int_C(x^2-y)\,ds&=2\int_0^\frac\pi4(2(1-\cos2t)-2\cos t)\,dt\\&=4\int_0^\frac\pi4(1-\cos2t-2\cos t)\,dt\\&=4\left[t-\frac12\sin2t-2\sin t\right]_0^\frac\pi4\\&=4\left(\frac\pi4-\frac12\sin\frac\pi2-2\sin\frac\pi4\right)\\&=4\left(\frac\pi4-\frac12-\sqrt2\right)\\&=\pi-2-4\sqrt2\end{align*}$$

Answer 5

thanks bro

Answer 6

no problem... I made a tiny error in simplfying though

Answer 7

The integrand should've been \(1-\cos2t-\cos t\) after pulling out another \(2\) to make the outside constant \(4\)... because of that we should actually have:\begin{align*}\int_C(x^2-y)\,ds&=2\int_0^\frac\pi4(2(1-\cos2t)-2\cos t)\,dt\\&=4\int_0^\frac\pi4(1-\cos2t-\cos t)\,dt\\&=4\left[t-\frac12\sin2t-\sin t\right]_0^\frac\pi4\\&=4\left(\frac\pi4-\frac12\sin\frac\pi2-\sin\frac\pi4\right)\\&=4\left(\frac\pi4-\frac12-\frac{\sqrt2}2\right)\\&=\pi-2-2\sqrt2\end{align*}

Answer 8

@rt212jsum apart from becoming my fan you may wish to give answers which helped you a medal :-p