# Finding a mass of elipse

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.

• "Elliptic curve" is a technical term in higher mathematics; it does not mean the same thing as "ellipse". The perimeter of an ellipse leads to what are known as "elliptic integrals", another advanced concept. Sep 17, 2013 at 13:25
• edited. Ok, I'm reading now about elliptic integrals. Sep 17, 2013 at 13:30
• Hm, I've read wiki about elliptic integrals, and I am quite sure I don't have to learn this concept to complete my task, because we don't have it in our program. There must be another way. Sep 17, 2013 at 13:41
• Well, are you sure you want to find the perimeter of an ellipse? You write of "mass" and "density", neither of which are relevant to finding the perimeter. Sep 17, 2013 at 13:44
• @GerryMyerson Oops, I have edited the second integral so that I can read it ;-p Sep 17, 2013 at 13:53

Here are some remarks:

• I think your integral should go from $\frac{3\pi}{2} \leq t \leq 2\pi$ if I'm not mistaken.
• $p\left(x(t),y(t)\right)$ simplifies to

$p\left(x(t),y(t)\right) = -7cos(t)sin(t)$

• Now substitute:

$u = sin^2(t) + 49 cos^2(t),$

$\frac{du}{dt} = 2(1-49) * cos(t)sin(t)$

$m= \int -7 cos(t)sin(t) * \sqrt{u} \frac{du}{2(1-49) cos(t)sin(t)}$

where the $cos(t)sin(t)$ cancel out. Is the rest clear?

• You're quite right about 3pi/2<=t<=2pi. But I'm not sure about simplification p(x(t),y(t))= -7cos(t)sin(t). Where did you lose/got rid of 2cost-7sint+2? Sep 17, 2013 at 16:25
• Do you agree with $x(t)=cos(t)-1$, $y=7 sin(t) + 2$? From there it's just writing all the terms and deleting the ones that cancel out (I double checked and still get the same result)
– Lisa
Sep 17, 2013 at 16:32
• I added the next step (solving the integral using the substitution rule)
– Lisa
Sep 17, 2013 at 16:40
• oh. I see, we made different substitutions. I brought ellipse to canonical form, assuming that it wouldn't change a curve length. But we still have eleptical integral. i'm quite sure the good idea would be to use polar cordinates here. So that what I'm trying now. Sep 17, 2013 at 16:40
• hm, I'll try your way. Sep 17, 2013 at 16:42