What are the correct ways to use polar form in solving integrals?

[transgalactic]

i added my questions and how i tried to solve them in the links bellow


http://img141.imageshack.us/my.php?image=img8210bs2.jpg

http://img81.imageshack.us/my.php?image=img8211hu3.jpg

 

[HallsofIvy]

For the first problem, find the length of the "astroid", x^2/3+ y<sub>2/3[/sup]= a^2/3, 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 x^2/3+ y^2/3= a^2/3sin²(t)+ a<sup>2/3cos2(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).</sup></sub>