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Homework Help: 2 question in integrals

  1. Jan 23, 2008 #1
  2. jcsd
  3. Jan 23, 2008 #2


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    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(([itex]\alpha[/itex]), y= a sin([itex]\alpha[/itex]). That is NOT "polar form. Polar form would be x= r cos([itex]\alpha[/itex]), y= r sin([itex]\alpha[/itex]), where both r and [itex]\alpha[/itex] are independent variables. There is no reason to thing that x= a cos(([itex]\alpha[/itex]), y= a sin([itex]\alpha[/itex]) 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= [itex]\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[/itex].

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

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