Finding probability of bivariate random variable from joint probability density function I have a Bivariate continuous random variable (X,Y) with joint probability density function
$ f_{X,Y}(x,y) = \{{6(y^2-x^2)}$ for $ 0 < x < y < 1$  (0 otherwise)
I want to work out $P(X + Y > 1) $
I have done $$\int_{1/2}^{1} \int_{1-y}^{1} 6(y^2-x^2) dx dy$$
$$ = \int_{1/2}^{1} (4y^3+6y^2-6y)dy$$
$$ = \frac{7}{16} $$
Is this right or have I made a mistake?
 A: The OP and the suggestion in the comment by @Zhanxiong are both wrong.
Draw a diagram of the non-zero density region 0 < x < y < 1 and of the region x + y > 1. It will be the box with vertices (0,0),(0,1),(1,0),1,1) with the non-zero density region being the northwest triangle formed in the box from the diagonal line y = x.  This must be intersected with the region x + y > 1 which is the northeast triangle formed in the box from the diagonal line x + y = 1.  Intersecting these regions is thus the upper triangle formed from the two diagonal lines.  
The OP is missing part of this upper triangle by integrating x from 0 to 1, which given y going from 1/'2 to 1 is equivalent to integrating x from 0 to 1/2, and therefore missing the term from the right-hand side of the upper triangle, i.e., for x > 1/2.  The correct integration is for x from 1-y to y and y from 1/2 to 1.  Or equivalently, form an additive term for each half of the upper triangle; the left half being for x from 1-y to 1/2 and y from 1/2 to 1, and the right half being for x from 1/2 to y and y from 1/2 to 1. Either way produces the answer of 5/8 (OP missed the 3/16 contribution from the right half of the upper triangle.
