Double integral on triangle using polar coordinates

sxyqwerty
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Homework Statement


Let R be the triangle defined by -xtanα≤y≤xtanα and x≤1 where α is an acute angle sketch the triangle and calculate
∫∫R (x2+y2)dA using polar coordinates
hint: the substitution u=tanθ may help you evaluate the integral

Homework Equations

The Attempt at a Solution


so the triangle has points (0,0) (1, xtanα) (1, -xtanα)
and r=1/cosα=secα
and I am stuck from here i don't know how to find the θ value of the polar coordinate
 
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The polar angle does not have one value. Like the radial coordinate r, it is an integration variable and you must integrate over it. The question you need to answer first is: What is the integral you need to solve and what are the integration boundaries?
 
i got ∫∫R r3drdθ i get that the boundary for r is 0 to secα but I am stuck after this...
i get that the polar angle is supposed to be multiple values (i.e. in a general form like θ or α) but i seriously have no clue on how to approach this further

from the original set of data i get ∫01-xtanαxtanαx2+y2dA
 
Last edited:
sxyqwerty said:
i get that the boundary for r is 0 to secα

This is wrong. I suggest you draw the triangle on a piece of paper and try to figure out which limits your integration variables have. Think about what the angles ##\alpha## and ##\theta## represent.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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