# Triple Integral

## Homework Statement

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

## 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.

## The Attempt at a Solution

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

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

1 person
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.

vanhees71
Gold Member
Hm, it says $x^2<1$. So what does that mean for the range of $x$?

1 person
Simon Bridge
Homework Helper
"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?

"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).

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 :)