Obtaining the probability based on two iid uniform Variable Let X and Y be two independent uniform random variables. Then, find
$P(\frac{1}{4} \leq X^2+Y^2\leq1)$.
I am not able to frame any idea about how this problem can be solved. One thing I know is that the probability region is a circle but not able to start. Also, the joint density will also be uniform. I have tried to write the expression in polar form but not sure it is correct. 
$\int_{0}^{2\pi}\int_{\frac{1}{4}}^{1}rdrd\theta  $
I am little doubtful whether the above expression is correct or not. 
Any help?
 A: Comment: One possible approach:
There are many uniform distributions. I assume you mean that $X$ and $Y$ are
independently distributed as $\mathsf{Unif}(0, 1),$ so that $(X, Y)$ is
uniformly distributed on the square with vertices at $(0,0)$ and $(1,1).$
From there, this can be regarded as a geometry problem, not necessarily a
calculus problem, following from @RichoKicked800goals's Comment. What proportion of the area of the square lies within
the region described by $1/4 < X^2 + Y^2 < 1?$
Below I used simulation as an easy way to make a plot, highlighting the 
target region in green. 

In case you want it, here is R code to make the figure:
set.seed(427)
x = runif(10^5); y = runif(10^5)
plot(x,y, pch=".")
cond = x^2 + y^2 < 1 & x^2 + y^2 > 1/4
points(x[cond], y[cond], col="green2", pch=".")
mean(cond);  pi*(1^2-.5^2)/4
## 0.58679    # fraction of green points
## 0.5890486  # area of green region

Note: If $X$ and $Y$ are independently distributed as $\mathsf{Unif}(-1,1),$
then the joint distribution has the square with vertices $(-1,-1)$ and $(1,1)$ as support, so that a sketch will show entire circles. But then the
density function is $f_{X,Y}(x,y) = 1/4$ over the support and $0$ elsewhere. (You'd need a factor $1/4$ inside your integral.)
