# 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)

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It would be better if you showed us your working. –  wj32 Oct 24 '12 at 22:23
I will try my best, although I don't know the coding to make it look nice. It will take a few minutes... –  Jared Oct 24 '12 at 22:26

You could start with letting $$u = \sqrt {{x^2} + {y^2}}$$ so that $$4\int {xy\sqrt {{x^2} + {y^2}} dy = \frac{4}{3}x{{({x^2} + {y^2})}^{3/2}}}.$$
Not sure what you mean. Note that $du = (1/2){({x^2} + {y^2})^{ - 1/2}} \cdot 2ydy = y{({x^2} + {y^2})^{ - 1/2}}dy$ so $dy = ({({x^2} + {y^2})^{1/2}}/y)du = (u/y)du$. Then $y$ in $dy$ and $y$ in the integrand cancel leaving $4x{u^2}$ as a new integrand. Remember that $x$ acts as a constant. –  glebovg Oct 24 '12 at 23:40