Convert triangle vertices to double integral polar coordiantes

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Homework Help Overview

The problem involves evaluating a double integral of the function f(x,y) = sqrt(x^2+y^2) over a triangular region defined by the vertices (0,0), (0,sqrt2), and (sqrt2,sqrt2). The original poster attempts to convert the integral into polar coordinates.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster discusses the conversion to polar coordinates and proposes specific bounds for the integral. Some participants question the correctness of these bounds and suggest alternative limits for theta based on geometric reasoning.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the bounds for theta and radius in the context of the triangular region. There is a suggestion that the bounds for theta may need to be adjusted, but no consensus has been reached.

Contextual Notes

Participants are considering the geometric implications of the triangle and how the lines intersect, particularly focusing on the angles formed by the lines y=x and the axes. There is an acknowledgment of the complexity in determining the correct bounds for the integral.

ramses07
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Homework Statement



integrate

f(x,y) = sqrt(x^2+y^2)

over triangle with vertices (0,0) (0,sqrt2) (sqrt 2, sqrt 2)

Homework Equations



x= rcosO, y = rsinO

x^2+y^2=r^2

The Attempt at a Solution



im supposed to use a double integral converted to polar coordinates,
so i used the bounds int. 0 to pi/4 int. 0 to sqrt 2 sec (r^2) drdO

are these the correct bounds? because i can't seem to find the answer.
 
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I think your bounds for theta should be pi/4 to pi/2
 
how do you know what the bounds are?
 
because the line y=x cuts it at a 45 degree angle and x=0 goes up to 90 so it goes from 45 to 90
like cutting a wedge out of a circle , but the radius is different , i think your bounds for the radius are correct.
 

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