A surface $$S(x,y)=2x+5y−3$$ is integrated once over a path consisting of the points that satisfy $$(x+1)^2+(y−1)^2=2$$; then what will be the value of this integral?
My Approach:
Required Integral is $$\oint_c S \cdot dR$$
where $\quad c : (x+1)^2 + (y-1)^2=2$
$dR=dx+dy+dz$
now parametric equation of $c$ is
$$x=(\surd 2) cos \theta - 1 \quad, y=(\surd 2) sin \theta + 1$$ $$\therefore I= \int_0^{2\pi} [2((\surd 2) cos \theta - 1)+5((\surd 2) sin \theta + 1)-3]\times[-(\surd 2) sin \theta d\theta+ (\surd 2) cos \theta d\theta]$$ $$\implies I=-6\pi + \int_{z_1}^{z_2} S \cdot dz$$
now how will I calculate $\int_{z_1}^{z_2} S \cdot dz$? Any Hints or suggestions please....
There is considerable ambiguity in this question. One possibility is that the integration is over a surface $\vec r=\langle x,y,z\rangle=\langle x,y,2x+5y-3\rangle$, We want to follow a path such that $(x+1)^2+(y-1)^2=2$ and we can readily parameterize the path along the surface as $$\vec r=\langle\sqrt2\cos\theta-1,\sqrt2\sin\theta+1,2\sqrt2\cos\theta-2+5\sqrt2\sin\theta+5-3\rangle$$ So $$d\vec r=\langle-\sqrt2\sin\theta,\sqrt2\cos\theta,-2\sqrt2\sin\theta+5\sqrt2\cos\theta\rangle d\theta$$ Then the element of arc length is $$ds=\left|\left|d\vec r\right|\right|=\sqrt{2\sin^2\theta+2\cos^2\theta+8\sin^2\theta-40\sin\theta\cos\theta+50\cos^2\theta}d\theta$$ So the arc length would be $$\begin{align}\oint ds&=\int_0^{2\pi}\sqrt{52\cos^2\theta-40\sin\theta\cos\theta+10\sin^2\theta}d\theta\\ &=\int_0^{2\pi}\sqrt{26(1+\cos2\theta)-20\sin2\theta+5(1-\cos2\theta)}d\theta\\ &=\int_0^{2\pi}\sqrt{31-20\sin2\theta+21\cos2\theta}d\theta\\ &=\int_0^{2\pi}\sqrt{31+29\cos(2\theta+\delta)}d\theta\\ &=\int_0^{2\pi}\sqrt{31+29(1-2\sin^2(\theta+\delta/2))}d\theta\\ &=\int_0^{2\pi}\sqrt{60-58\sin^2\theta}d\theta\\ &=8\sqrt{15}\int_0^{\frac{\pi}2}\sqrt{1-\frac{29}{30}\sin^2\theta}d\theta\end{align}$$ And that's a complete elliptic integral of the second kind. In the above, $\tan\delta=20/21$.
Another interpretation is that the integration takes place on the $xy$-plane so that $$\vec r=\langle\sqrt2\cos\theta-1,\sqrt2\sin\theta+1\rangle$$ Then $$d\vec r=\langle-\sqrt2\sin\theta,\sqrt2\cos\theta\rangle d\theta$$ Now there are two subinterpretations. The first is that you want to integrate $$\oint S(x,y)d\vec r$$ Which would be $$\begin{align}\vec I&=\int_0^{2\pi}(2\sqrt2\cos\theta+5\sqrt2\sin\theta)\langle-\sqrt2\sin\theta,\sqrt2\cos\theta\rangle d\theta\\ &=\int_0^{2\pi}\langle-4\sin\theta\cos\theta-10\sin^2\theta,4\cos^2\theta+10\sin\theta\cos\theta\rangle d\theta\\ &=\langle-4(0)-10(\pi),4(\pi)+10(0)\rangle\\ &=\langle-10\pi,4\pi\rangle\end{align}$$ Or finally you might be asking for $$\oint S(x,y)ds$$ So you need the scalar element of arc length $$ds=\left|\left|d\vec r\right|\right|=\sqrt2d\theta$$ And then we arrive at $$I=\int_0^{2\pi}(2\sqrt2\cos\theta+5\sqrt2\sin\theta)\sqrt2d\theta=0$$ Hopefully one of the above is the answer you've been struggling with for the last month or so.