(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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]

3. 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]I found one mistakes in letter

b)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

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: 2 exercises on changing variables of double integrals

**Physics Forums | Science Articles, Homework Help, Discussion**