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

In summary, the conversation is discussing the evaluation of an integral with the range and limits of integration given as 0<=z<=y<=x^2<=1. The attempt at a solution includes an integral with different limits, but the correct answer should be (1/3)*(13/15 - pi/4). The discussion also touches on the meaning of the notation and clarifies the range for x as being from -1 to 1.
  • #1
Swasse
9
0

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.
 
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  • #2
The notation does not seem to make any sense in this context.
The only thing I can think of is:

$$\int_0^1 dx \int_0^{x^2} dy \int_0^y dz \frac{z}{1+x^2}$$
 
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  • #3
Simon Bridge said:
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!
 
  • #4
Hm, it says [itex]x^2<1[/itex]. So what does that mean for the range of [itex]x[/itex]?
 
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  • #5
"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?
 
  • #6
Simon Bridge said:
"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).
 
  • #7
vanhees71 said:
Hm, it says [itex]x^2<1[/itex]. So what does that mean for the range of [itex]x[/itex]?

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

1. What is a triple integral?

A triple integral is a mathematical tool used to calculate the volume under a three-dimensional surface. It is similar to a regular integral, which calculates the area under a curve, but the triple integral takes into account three dimensions (x, y, and z) instead of just two.

2. How do I interpret the given bounds for this triple integral?

In this specific problem, the bounds are 0 <= z <= y <= x^2 <= 1. This means that the integral is being calculated over a region where z is between 0 and 1, y is between 0 and x^2, and x is between 0 and 1. This can be visualized as a three-dimensional shape with these boundaries.

3. What is the process for solving a triple integral?

To solve a triple integral, you need to first determine the bounds and the function being integrated. Then, you need to set up the integral using the appropriate notation (either Cartesian, cylindrical, or spherical coordinates). Finally, you can use various techniques such as change of variables or symmetry to simplify the integral and solve it using standard integration methods.

4. Can this triple integral be solved using any coordinate system?

Yes, triple integrals can be solved using different coordinate systems such as Cartesian, cylindrical, and spherical coordinates. The choice of coordinate system often depends on the given boundaries and the complexity of the function being integrated.

5. How can I check if my solution to a triple integral is correct?

You can check your solution by using a computer program or calculator to evaluate the integral. You can also use visualization tools to plot the three-dimensional shape and see if your calculated volume matches the actual shape. Additionally, you can compare your solution to a known solution or ask a colleague or instructor to review your work.

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