Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite
2

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)

share|improve this question
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... – Izzy Ward Oct 24 '12 at 22:26

1 Answer

up vote 1 down vote accepted

You could start with letting $$u = \sqrt {{x^2} + {y^2}}$$ so that $$4\int {xy\sqrt {{x^2} + {y^2}} dy = \frac{4}{3}x{{({x^2} + {y^2})}^{3/2}}}.$$

share|improve this answer
But that would make du = 1/2 (x^2 + y^2)^(-1/2) * 2y = y/(x^2+y^2)^(1/2). That is not in the original integral, so it is not possible to do without an extraneous amount of work to balance that out (at least not for my level of training). – Izzy Ward Oct 24 '12 at 23:25
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
Wow that is beyond the scope of what we have covered in my Calc 3 class but now I understand what you're getting at. I'll try the problem with your strategy. Thanks! – Izzy Ward Oct 24 '12 at 23:43
You could try polar coordinates. Have you covered that? – glebovg Oct 24 '12 at 23:51
1  
Go to wolframalpha.com. WolframAlpha is very user-friendly. Simply search for: double integral of 4*x*y*sqrt(x^2+y^2) from x = 0 to x = 3 and y = 0 to y = 1. See Examples if you are having trouble. – glebovg Oct 25 '12 at 0:29
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.