- #1
timmastny
- 10
- 0
Homework Statement
Given [itex]\int^{\sqrt{6}}_{0}[/itex][itex]\int^{x}_{-x}[/itex]dydx, convert to ploar coordinates and evaluate.
Homework Equations
We know that x=rcos[itex]\theta[/itex] and y=rsin[itex]\theta[/itex] and r =x^2+y^2
The Attempt at a Solution
First, I defined the region of the original integral: R = 0 [itex]\leq[/itex]x[itex]\leq[/itex] [itex]\sqrt{6}[/itex] and -x[itex]\leq[/itex]y[itex]\leq[/itex]x.
In other words, we know that y is bounded on the lower limit by the line -x and the upper limit x. Because the line x produces a 45 degree angle, -[itex]\pi[/itex]/4 [itex]\leq[/itex] [itex]\theta[/itex] [itex]\leq[/itex] [itex]\pi[/itex]/4.
However, I do not understand how to find domain of r. The region the original integral covers is not circular. We can prove this because at ([itex]\sqrt{6}[/itex],0), r = [itex]\sqrt{6}[/itex] and at ([itex]\sqrt{6}[/itex], [itex]\sqrt{6}[/itex]), which is still in the region defined by the original integral, r = [itex]\sqrt{12}[/itex] ([itex]\sqrt{\sqrt{6}^2+\sqrt{6}^2}[/itex])
So that is where I am stuck. The answer to the problem according to the book is 6 but I do not know how to get there. All the examples we worked were circular regions.
Additionally, assuming the region is 0 [itex]\leq[/itex] r [itex]\leq[/itex] [itex]\sqrt{6}[/itex] does not work either. [itex]\int^{pi/4}_{-pi/4}[/itex] [itex]\int^{sqrt{6}}_{0}[/itex] r dr d[itex]\theta[/itex] = (3[itex]\pi[/itex])/2
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