# Line integration over a 3-dimensional curve

Consider a curve C that is defined by the intersection of the following surfaces: $$x+y+z=0$$ $$and$$ $$x^2+y^2+z^2=K$$ for some non-zero, positive real number K.

Find $$I=\int_Cy^2ds$$

I first tried to find the equation for the curve in terms of a vector r(t) and I obtained: $$r(t)=\left\langle \frac{\cos t}{\sqrt{ 2+\sin 2t}},\frac{\sin t}{\sqrt{ 2+\sin 2t}},\frac{-\cos t - \sin t}{\sqrt{ 2+\sin 2t}}\right\rangle$$ But then, to solve for I with this method, one must take the magnitude of the derivative of r, which becomes incredibly complicated. My professor claims there is an easier way to solve this problem, but I cannot figure anything out.

• Do you miss the $K$ somewhere in $r(t)$? – Jack Dec 3 at 1:07
• This is the intersection of a sphere with a plane, i.e. a circle. There are much simpler parametrizations of a circle than yours, namely linear in sines and cosines, see Parametric Equation of a Circle in 3D Space?. – Conifold Dec 3 at 1:45
• Consider that $\int_C x^2 ds = \int_C y^2 ds = \int_C z^2 ds$. – Maxim Dec 3 at 4:47

Here is a „non brute-force“ approach (assume WLOG $$K=1$$.)

Note that, by symmetry of the curve, we have $$\int_C x^2 \,\mathrm ds= \int_C y^2 \,\mathrm ds =\int_C z^2 \,\mathrm ds.$$

Hence, $$3 \int_C y^2 \,\mathrm ds=\int_C x^2+y^2+z^2\,\mathrm ds = \int_C 1\,\mathrm ds=\text{Length of the curve } c.$$

It is known that the intersection of a sphere with a plane is a circle (under certain conditions which are fulfilled here.)

Also, as can be seen also from your parametrization (compute the distance of two points with time $$\pi$$ apart), our circle has a diameter of $$2$$.

So the length of the curve $$c$$ is simply the circumference of a circle with radius $$2$$ which is $$2\pi$$.

So, by our above argument, $$\bbox[5px,border:2px solid #C0A000]{ I=\int_C y^2\,\mathrm ds = \frac{\text{Length of }c}3=\frac{2\pi}3.}$$

Here is a „brute-force approach“ (assume WLOG $$K=1$$.)

Even with your parametrization the calculations are doable:

Note that $$r'(t)=\left(-\frac{2 \sin (t)+\cos (t)}{(\sin (2 t)+2)^{3/2}},\frac{\sin (t)+2 \cos (t)}{(\sin (2 t)+2)^{3/2}},\frac{\sin (t)-\cos (t)}{(\sin (2 t)+2)^{3/2}}\right)$$

so that $$\|r'(t)\|_2=\sqrt{\frac{(\sin (t)-\cos (t))^2}{(\sin (2 t)+2)^3}+\frac{(\sin (t)+2 \cos (t))^2}{(\sin (2 t)+2)^3}+\frac{(2 \sin (t)+\cos (t))^2}{(\sin (2 t)+2)^3}}$$

which reduces to $$\|r'(t)\|_2=\sqrt{\frac{6+6\sin(t)\cos(t)}{(2+\sin(2t))^3}}=\sqrt{3} \sqrt{\frac{1}{(\sin (2 t)+2)^2}}.$$

So $$I=\int_0^{2\pi} \frac{\sqrt{3} \sin ^2(t)}{(\sin (2 t)+2)^2}\,\mathrm dt=2\sqrt{3}\int_0^{\pi} \frac{\sin ^2(t)}{(\sin (2 t)+2)^2}\,\mathrm dt.$$

It remains to calculate the last integral, which I will call $$J$$. $$J$$ is a standard integral and we may proceed as follows: $$J=\int_0^\pi \frac{\sin^2(t)}{(2\sin(t)\cos(t)+1)^2}\,\mathrm dt\overset{(1)}=\int_0^\pi \frac{\csc^2(t)}{4\cot^2(t)+4\csc^4(t)+8\cot(t)\csc^2(t)}\,\mathrm dt\overset{(2)}=\int_{-\infty}^\infty\frac{1}{4(u^2+u+1)^2}\,\mathrm du.$$

(1): Expand the fraction by $$\csc^4$$
(2): Use that $$\csc^2=\cot^2+1$$ and substitute $$u=\cot(t)$$.

So, by substituting $$s=u+\frac12$$, $$J=\frac14\int_{-\infty}^\infty \frac{1}{(s^2+\frac34)^2}\,\mathrm ds=\frac{4}9\int_{-\infty}^\infty \frac{1}{(\frac43 s^2+1)^2}\,\mathrm ds.$$

The last integral can be evaluated as here.

The final result should be $$J=\frac{\pi}{3\sqrt 3}$$ so that $$\bbox[5px,border:2px solid #C0A000]{I=\frac{2\pi}3.}$$