Triple Integral Homework: Solve 0<=z<=y<=x^2<=1

[Swasse]

Homework Statement

∫∫∫z/(1+x^2)dxdydz with the range 0<=z<=y<=x^2<=1

Homework Equations

The Attempt at a Solution

I have tried with the limits 0<=z<=1; z<=y<=1 and sqrt(y)<=x<=1, but it doesn't get me the right answer. Can you please help me and maybe give me a step-by-step solution.

 

[Simon Bridge]

The notation does not seem to make any sense in this context.
The only thing I can think of is:

$$\int_0^1 dx \int_0^{x^2} dy \int_0^y dz \frac{z}{1+x^2}$$

 

[Swasse]

The notation does not seem to make any sense in this context.
The only thing I can think of is:

$$\int_0^1 dx \int_0^{x^2} dy \int_0^y dz \frac{z}{1+x^2}$$

What I meant with 0<=z<=y<=x^2<=1, was that the area of integration is D, where D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. Does that make more sense?

Anyway, I tried to evaluate your integral and got (1/6)*(13/15 - pi/4), but it should be twice as much according to my key. Your limits seems correct, so I don't know where the error is. Can someone shed some light on this.

Thanks for your reply by the way!

 

[vanhees71]

Hm, it says $x^2<1$. So what does that mean for the range of $x$?

 

[Simon Bridge]

"area of integration"?
Suggests the limits are not as right as they look - maybe vanhhees is on to something?
Perhaps you should tell us what the integral is supposed to find?
Are you trying to find a volume?

 

[Swasse]

"area of integration"?
Suggests the limits are not as right as they look - maybe vanhhees is on to something?
Perhaps you should tell us what the integral is supposed to find?
Are you trying to find a volume?

Range or volume of integration might be more correct, English is not my first language so I hope you understand anyway. The problem just says to evaluate the integral over the set D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. The answer should be (1/3)*(13/15 - pi/4).

 

[Swasse]

Hm, it says $x^2<1$. So what does that mean for the range of $x$?

So the range for x should be from -1 to 1? That makes sense :)