I have a task, to find a part of perimeter of an ellipse (on plane), I know it's density function. Both ellipse equation and density are given in cartesian coordinates. So I set up a line integral, but it seems to me quite complex to calculate: $${1\over4}\int_0^{2\pi}(2\cos t-7\sin t-\cos t\sin t+2)\sqrt{(7\cos t)^2+(-\sin t)^2}\,dt$$ or integral $$\int_{\frac{\pi}{2}}^{\pi}(2\cos t-7\sin t-\cos t\sin t+2)\sqrt{(7\cos t)^2+(-\sin t)^2}\,dt$$ both integrals should give the same result. How can I solve this? I thought about trying to rewrite all in polar coordinates, but got confused, because I'm not quite sure how.
Find a mass of a line arc
$$(x+1)^3 + {(y-2)^2\over49}=1, x>=-1, y<=2$$. Density: $$p(x,y)=2x-y-xy+2$$.
What was my logic: $$m=\int_{L}p(x,y)dl$$ where p - density, m - mass, L- is a quarter of ellipse perimeter length curve.
Here are some remarks:
$p\left(x(t),y(t)\right)$ simplifies to
$p\left(x(t),y(t)\right) = -7cos(t)sin(t) $
Now substitute:
$u = sin^2(t) + 49 cos^2(t),$
$\frac{du}{dt} = 2(1-49) * cos(t)sin(t)$
leads to
$m= \int -7 cos(t)sin(t) * \sqrt{u} \frac{du}{2(1-49) cos(t)sin(t)} $
where the $cos(t)sin(t)$ cancel out. Is the rest clear?