A wire has the shape of a curve obtained by the intersection of the portion of the sphere $x^2+y^2+z^2=4$, $y\geq 0$, with the plane $x+z=2$. Knowing that the density in each point of the wire is given by $f(x,y,z)=xy$, evaluate the total mass of the wire.
Well, I found that the curve is given by $$\alpha(t)=\left(t,\sqrt{4t-2t^2},2-t\right),$$ but the integral was a bit ugly and I couldn't solve. So, I need some help.
On
You can change variables twice, first with $t = 1+u$, then with $v=cos(u)$ :
$$ \begin{align*} \int{ t \sqrt{4t-2t^2}}\text{d}t &= \int{ t \sqrt{2t(2-t)}}\text{d}t\ \\ &= \int (1+u)\sqrt{2-2u^2}\text{d}u \\ &= \sqrt{2}\int \sqrt{(1-cos^2(v))}(1+cos(v))\text{d}v \end{align*}$$
Using basic trigonometric formulas ($sin^2v=1-cos^2v$ and $sin(v) cos(v) = \frac{1}{2}sin(2v)$), you can finish the calculation.
Note that $$|\alpha'(t)|^2=\left|\left(1,\frac{d}{dt}\sqrt{4t-2t^2},-1\right)\right|^2=1+\frac{2(1-t)^2}{t(2-t)}+1=\frac{2}{{t(2-t)}}.$$ Hence the total mass of the wire can be evaluate by the following line integral \begin{align*} \int_{t=0}^2x(t)y(t)|\alpha'(t)|\,dt=\sqrt{2}\int_0^2t\sqrt{t(2-t)}\,\frac{\sqrt{2}}{\sqrt{t(2-t)}}\,dt=2\int_0^2 tdt=4. \end{align*}