Cartesian to Polar Integral: Evaluate

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Homework Help Overview

The discussion revolves around converting a Cartesian integral into its polar form and evaluating it. The integral in question involves the expression for the area under a curve defined by the limits of integration in Cartesian coordinates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of the integral and the appropriate limits for polar coordinates. There is uncertainty regarding the new boundaries for integration and the interpretation of the original limits.

Discussion Status

Some participants have provided insights into the relationship between Cartesian and polar coordinates, while others are questioning the validity of the integration boundaries. There is an ongoing exploration of how to accurately represent the integration region in polar coordinates.

Contextual Notes

Participants note the need for a visual representation of the integration region to clarify the boundaries for r and theta. There is also mention of potential misunderstandings regarding the limits derived from the original Cartesian setup.

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
 
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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.
 
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?
 
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.
 
i thought it was x=sqrt(1-y)
 

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