There is a continuous integrable function f(x,y,z)=x 2y 2z 3 in three-dimensional space, which requires the calculation of its triple integral in the three-dimensional region v enclosed by three coordinate planes x=0, y=0 and z=1 and a rotating parabola x 2y 2=z.
Solution: Since this problem involves a function defined in three-dimensional space and a complex three-dimensional region, we need to use triple integration to solve it. For this type of integration, it is crucial to choose the right order of integration, which will directly affect the ease of calculation. In this problem, we can fix the z-variable first, then double integrate the region on the xy plane, and then integrate z.
1. Determine the integral range:
For z, integrate from 0 to 1 because the stereoregional region is defined by the z=0 plane and the z=1 plane.
For x and y, at each fixed value of z, they satisfy the condition of x 2y 2=z, which is actually on a circle with the origin as the center and the radius z. We need to integrate this circle.
2. Set the integration order and convert the coordinates
To simplify the integration process, we can use polar transformations. Let x=rcos , y=rsin, then there is r 2=z, so 0 r z, 0 2.
We first integrate from 0 to 2, then from 0 to z for r, and then from 0 to 1 for z.
3. Expression conversion and integral calculation
Convert f(x,y,z)=x 2y 2z 3 to polar coordinates to give f(r, ,z)=r 2(cos 2 sin 2 )z 3=r 2z 3.
Thus, the original triple integral can be expressed as:
From 0 to 1) (from 0 to z) (from 0 to 2) r 2z 3rdrd dz
Integrate first, then r, and then z, and get the final result.
The specific integration steps are as follows:
i= (from 0 to 1)z3( (from 0 to z)r 3dr)( from 0 to 2)d )dz
From 0 to 1)Z3(R44)|0^(√z)(2π)dz
(from 0 to 1) z 3 (z 2 4) dz
4 (from 0 to 1) z 5dz
4z^6/6|0^1
The function f(x,y,z) has a triple integral value of 24 in a given solid region v.