Triple Integral Evaluation for Bounded Region with Polynomial Boundaries

iamalexalright
Messages
157
Reaction score
0

Homework Statement


Evaluate triple integral of 3xy over the bounded region:

y = x^{2}
x = y^{2}
z = 6x + y


The Attempt at a Solution


Bounds on integral would be:
0 \leq x \leq 1
x^{2} \leq y \leq \sqrt{x}
0 \leq z \leq 6x + y

Correct?
 
Physics news on Phys.org
If they also gave you the bound z>=0 that looks fine. Otherwise the region isn't bounded.
 
Yep, z = 0 included

I get 9/8. Btw, what do you use to calculate them ?
 
iamalexalright said:
Yep, z = 0 included

I get 9/8. Btw, what do you use to calculate them ?

I use a software package called Maxima to actually do the integrals. I use my head to get the limits. What do you use? And I do get 9/8. We seem to be getting lucky here. I've usually made a mistake by now.
 
Just started using wolframalpha.com for the integrals. Just wanted to make sure my syntax and what not was correct.

Have to draw it out to get the limits heh
 
iamalexalright said:
Just started using wolframalpha.com for the integrals. Just wanted to make sure my syntax and what not was correct.

Have to draw it out to get the limits heh

You must be doing it right. Sure, the integration part is mechanical, no reason not to use a calculator there. The head work is the hard part and you seem to be getting it right.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top