# 2 question in integrals

1. Jan 23, 2008

### transgalactic

2. Jan 23, 2008

### HallsofIvy

Staff Emeritus
For the first problem, find the length of the "astroid", x2/3+ y2/3[/sup]= a2/3, you say you wanted to put it in "polar form" and then give x= a cos(($\alpha$), y= a sin($\alpha$). That is NOT "polar form. Polar form would be x= r cos($\alpha$), y= r sin($\alpha$), where both r and $\alpha$ are independent variables. There is no reason to thing that x= a cos(($\alpha$), y= a sin($\alpha$) satisfy the equation of the curve.

As for the "method" your text uses, you say they let x= a sin3(t), y= a cos3(t). Okay, then x2/3+ y2/3= a2/3sin2(t)+ a2/3cos2(t)= a2/3 so those are parametric equations for the curve. Also, then dx= 3a cos(t)sin(2(t)dt and dy= -3a sin(t)cos2(t) dt so that ds= $\sqrt{(dx/dt)^2+ (dy/dt)^2}= \sqrt{9a^2 cos^2(t)sin^3(t)+ 9a^2 sin^2(t) cos^4(t)}dt$.

As for the second one, y2= x2(a2- x2, i don't see why you again have "x= a cos$\alpha$, y= a sin$\alpha$". Polar coordinates are x= r cos$\theta$, y= r sin[\theta]. (of course, it doesn't matter whether you use $\alpha$ or $\theta$. $\theta$ is the standard notation.)

Then $y^2= r^2 sin^2(\theta)$ and $x^2= r^2 cos^2(\theta)$ so your equation becomes $r^2 sin^2(\theta)= r^2(a^2- r^2 cos^2(\theta)$. We can cancel the two $r^2$ terms and then we have $sin^2(\theta)= a^2- r^2 cos^2(\theta)$ so $r^2 cos^2(\theta)= a^2- sin^2(\theta)$, $r^2= a^2 sec^2(\theta)- tan^2(\theta)$.