I've been doing some double integral questions out of the Glyn James, advanced modern engineering mathematics book and a particular question stumped me. Up until this point i had been OK going through the questions. I've now double and then triple checked and still can't see where I've gone wrong i don't have anyone to consult so thought i would ask here, the solution in the book is simply a numerical answer and at this point i'm even entertaining the idea the solution in the book is incorrect.
Here is the question:
$$\int_{0}^{1}\int_{0}^{\sqrt{x-x^2}}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big)dydx$$
The first thing i did was work out that $\pm\sqrt{x-x^2}$ is actually an unusual equation of a circle that is a circle of the form $(x-1/2)^2+y^2=(1/2)^2$ i then sketched this semi circle out as my region. Next i then changed coordinates to polar coordinates using the Jacobian noting that the parametric equation for x in this case is $x=\frac{1}{2}+rcos(\theta)$. This then gives us the following:
$$\int_{-}^{-}\int_{-}^{-}\Big(\frac{1}{2}+rcos\Big)drd\theta$$
Looking at the new limits, we see that the angle ranges in the interval $[0,\pi]$ and the r value (not sure about this) ranges in the interval $[0,1/2]$.
$$\int_{0}^{\pi}\int_{0}^{1/2}\Big(\frac{1}{2}+rcos\Big)drd\theta$$
Evaluating this integral i get $\frac{\pi}{4}$ the book however says it's $\frac{1}{3}$. Any pointers to why i'm wrong would be great i have some doubts mostly about the limits in particular the limits for r.
Best Regards
Best Answer
The polar equation of the region of integration is $r=\cos\theta$ with $r$ going from $0$ to $\cos\theta$ and $\theta$ going from $0$ to $\dfrac{\pi}{2}$.
So you want
$$ \int_0^{\pi/2}\int_0^{\cos\theta}(\cos\theta)\,rdr d\theta $$