Solve line integral $\int_\limits{C}x^2\;\mathrm{d}s$

Question

The integral:

$$\int_\limits{C}x^2\;\mathrm{d}s$$

where $C$ is a circle $x^2+y^2+z^2=a^2, x+y+z=0$

I do not know why 2 lines are given, and how should I find $\mathrm{d}s$ then

Answer 1

The two things given are a sphere and a plane, whose intersection yields a a circle.

$(0,1,-1)$ is orthogonal to $(1,1,1)$ and $(0,1,-1)\times(1,1,1)=(2,-1,-1)$ is orthogonal to both. Thus. $$ \left\{\frac{(1,1,1)}{\sqrt3},\frac{(0,1,-1)}{\sqrt2},\frac{(2,-1,-1)}{\sqrt6}\right\} $$ is an orthonormal basis for $\mathbb{R}^3$. Thus, a parametric formula for the circle is $$ (x,y,z)=a\,\frac{(0,1,-1)}{\sqrt2}\,\cos(t)+a\,\frac{(2,-1,-1)}{\sqrt6}\,\sin(t) $$ Thus, with $s=at$, we have $$ x=\frac{2a}{\sqrt6}\,\sin(t) $$ Therefore, $$ \begin{align} \int_0^{2\pi a} x^2\,\mathrm{d}s &=\frac{2a^3}3\int_0^{2\pi}\sin^2(t)\,\mathrm{d}t\\[6pt] &=\frac{2\pi a^3}3 \end{align} $$

Answer 2

Since $z=-x-y$, we obtain $$x^2+xy+y^2=\frac{a^2}{2}$$ or $$\frac{3}{4}x^2+\left(\frac{1}{2}x+y\right)^2=\frac{a^2}{2}.$$ Let $\frac{\sqrt{3}}{2}x=\frac{a}{\sqrt2}\cos{t}$ and $\frac{1}{2}x+y=\frac{a}{\sqrt2}\sin{t}$.

Thus, $x=a\sqrt{\frac{2}{3}}\cos{t}$, $y=a\left(\frac{1}{\sqrt2}\sin{t}-\frac{1}{\sqrt6}\cos{t}\right)$ and $z=a\left(-\frac{1}{\sqrt2}\sin{t}-\frac{1}{\sqrt6}\cos{t}\right)$. $$ds=\sqrt{(x')^2+(y')^2+(z')^2}dt=$$ $$=a\sqrt{\frac{2}{3}\sin^2t+\left(\frac{1}{\sqrt2}\cos{t}+\frac{1}{\sqrt6}\sin{t}\right)^2+\left(-\frac{1}{\sqrt2}\cos{t}+\frac{1}{\sqrt6}\sin{t}\right)^2}dt=$$ $$=a\sqrt{\sin^2t+\cos^2t}dt=adt.$$ Id est, $$\int\limits_{C}x^2ds=\frac{2}{3}a^3\int\limits_{0}^{2\pi}\cos^2tdt=\frac{2}{3}a^3\int\limits_{0}^{2\pi}\left(\frac{1}{2}+\frac{1}{2}\cos2t\right)dt=\frac{2\pi}{3}a^3.$$ Done!