Cartesian to polar

  • Thread starter chevy900ss
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  • #1
chevy900ss
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Homework Statement



Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
int(-1to1)int((sqrt(1-y^2))to(sqrt(1-y))[x^2+y^2]dxdy

Homework Equations


x=rcostheta
y=rsintheta


The Attempt at a Solution


int(-1to1)int((sqrt1-(rsintheta)^2)to(sqrt(1-(rsintheta))))[(rcostheta)^2+(rsintheta)^2]rdrdtheta
 

Answers and Replies

  • #2
CompuChip
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Note that [x^2+y^2] is simply r^2 (by definition!).

Also you cannot just change your integration boundaries like that. For example, neither theta nor r runs between -1 and 1.
I suggest drawing an image of the integration region, then think about how to describe it in terms of boundaries on r and theta.
 
  • #3
chevy900ss
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ok so i am not sure how to get the new boundries
but its r^2rdrdtheta which integrated goes to r^4/4dtheta. Is this right?
 
  • #4
HallsofIvy
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Yes, [itex]\int r^3dr[/itex] is [itex]r^4/4[/itex].

Are you sure about that "[itex]y= \sqrt{1- x}[/itex] limit? That will give the right part of a parabola for y> 0 it is inside the unit circle given by the lower limit and for y< 0, it is outside.
 
  • #5
chevy900ss
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i thought it was x=sqrt(1-y)
 

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