Statistics, joint probability density

If the independent random variables X and Y have the marginal densities
f(x) = {1/4, 0 < x < 4; 0, otherwise}
π(y) = {1/6, 0 < y < 6; 0, otherwise}
find
(a) the joint probability density of X and Y;
(b) the value of P(X^2 + Y^2 > 1).

Someone already said the answer to (a) is probably f(x,y) = 1/24, which seems to be fine.
[f(x)π(y) = (1/4)(1/6) = 1/24 = f(x,y); true iff. X and Y are independent]

I'm having trouble with (b), though. I figure I should perform double integration on f(x,y), but what are the bounds? Someone please tell me if I'm doing this wrong:
0 < x < 4 => 0 < x^2 < 16
0 < y < 6 => 0 < y^2 < 36
==>
0 < x^2 + y^2 < 52
But we want 1 < x^2 + y^2 < 52.
So √(1 - y^2) < x < √(52 - y^2)
and √(1 - x^2) < y < √(52 - x^2).
Then 1 < x < 4 and 1 < y < 6 should be the intervals we want, right?
So P(X^2 + Y^2 > 1) = P(1 < x < 4, 1 < y < 6)
= ∫[1 to 6] ∫[1 to 4] 1/24 dx dy
= ∫[1 to 6] 1/8 dy
= 5/8.
Right? Again, I'm pretty lost with all this, so please correct me if I'm wrong.

The trick for (b) is to use complements
P(X^2 + Y^2 > 1)
= 1 - P(X^2 + Y^2 ≤ 1)
= 1 - ∫∫ (1/24) dA, over the region X^2 + Y^2 ≤ 1
= 1 - (1/24) * (Area of unit circle)
= 1 - π/24.

I hope this helps!