- #1
carlosbgois
- 68
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Homework Statement
a) [itex]\int\int_{B}\frac{\sqrt[3]{y-x}}{1+y+x} dxdy[/itex], where B is the triangle with vertices [itex](0, 0), (1, 0), (0, 1)[/itex].
b)[itex]\int\int_{B}x dxdy[/itex] where B is the set, in the xy plane, limited by the cardioid [itex]ρ=1-cos(θ)[/itex]
The Attempt at a Solution
a) Let [itex]ψ: \left\{u = y-x, v = 1+y+x\right\}[/itex]. Then ψ has an inverse of class [itex]C^{1}[/itex], and the jacobian of this transform is [itex]\frac{-1}{2}[/itex], and also B is limited by [itex]x=0, y=0, y=-x+1[/itex], which over the ψ transform becomes, respectively [itex]v=u+1, v=-u-1, v=2[/itex]. For last, by drawing perpendicular u and v axis and limiting the region, I could see that the first two transforms intercept and [itex]v=0[/itex]. Then I did [itex]\frac{-1}{2}\int^{2}_{1}\left[\int^{v-1}_{1-v}\frac{\sqrt[3]{u}}{v} du\right]dv[/itex].
and I don't think this is going anywhere because most values of u inside the transformed B are negative, and the solution will have a complex part, which I'm sure it shouldn't.
b) (this attempt has a mistake, see edit below)As the cardiod is is polar coordinates, we may change x to those coordinates too, hence [itex]x=ρcos(θ)[/itex]. As the image of the cosine is [itex][-1, 1][/itex], then [itex]0≤ρ≤2[/itex], and I also made [itex]0≤θ≤2\pi[/itex], hence the integral becomes [itex]\int^{2\pi}_{0}\left[cos(θ)\int^{2}_{0}ρ^{2} dρ\right] dθ[/itex].
By solving this I got 0, and I don't think it's correct.
[EDIT]
b) I found one mistakes in letter b: the interval for ρ is not as said before, it is [itex]0≤ρ≤1-cos(θ)[/itex], hence the 'correct' integral is [itex]\frac{1}{3}\int^{2\pi}_{0}cos(θ) (1-cos(θ))^{3} dθ.[/itex]
For solving this I have expanded the integrand and determined the first two terms using integration by parts, and the other two were direct. My result is [itex]\frac{-5\pi}{4}[/itex].
New questions: Is the result correct? Is there any faster way to integrate the expression, instead of expandind and evaluating term by term?
Thanks
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