<s>[INST] Iterated Integral requiring u-substitution This is an iterated integral. I have tried solving it several times using u-substition, but I am not getting the correct answer. My latest result is (4/15)(10^(5/2)-33). Obviously something is off, but what? Could you please show me some steps and your final answer so I can work it out on my own and make sure I get it right?
Calculate the iterated integral:
$$
\int_0^3\int_0^1 4xy\sqrt{x^2+y^2}~dy~dx
$$
My latest attempt:
u=x^2+y^2  _  du=2ydy _ @y=1, u=x^2+1 _ @y=0, u=x^2
int(0 to 3)[int(x^2 to x^2+1) 2xu^(1/2)du]dx = int(0, 3)[4/3 * xu^(3/2) for u= x^2 to u=x^2+1]dx = 4/3 * int(0 to 3) [x(x^2+1)^(3/2) - x(x^2)^(3/2)]dx 
= 4/3 int(0 to 3) [x(x^2+1)^(3/2)]dx - 4/3 int(0 to 3) [x^4]dx
v= x^2+1 _ dv=2xdx _ @x=3, v=10 _ @x=0, v=1
4/3 int(1 to 10) [1/2 * v^(3/2)]dv - 4/3 int(0 to 3) [x^4]dx
= 2/3[2/5 * v^(5/2) for v=1 to v=10] - 4/3[1/5 * x^5 for x=0 to x=3]
=4/15(10^(5/2) - 1) - 4/15 (32 - 0) = 4/15(10^(5/2) - 33)
 [/INST] You could start with letting
$$u = \sqrt {{x^2} + {y^2}}$$
so that
$$4\int {xy\sqrt {{x^2} + {y^2}} dy = \frac{4}{3}x{{({x^2} + {y^2})}^{3/2}}}.$$
</s>