Solving a Triple Integral Problem with Effective Coordinate Transformation

Triple integrals are a fundamental tool in multidimensional calculus, often used to calculate volumes, masses, and other quantities in three-dimensional space. In this article, we will walk through solving a specific triple integral problem using effective coordinate transformation and the Jacobian determinant. This approach simplifies the integration process while maintaining accuracy.

Problem Statement

Consider the following triple integral problem:

( iiint_{Omega} x , dx , dy , dz )

The region ( Omega ) is located in the first octant of space, where all coordinates are positive. The region is bounded by the cylinder ( y^2 16x ), the plane ( frac{y}{4} frac{z}{7} 1 ), and the coordinate plane.

Solution Approach

To solve this problem, we will follow the suggested coordinate transformation and compute the Jacobian determinant. This approach allows us to avoid tedious calculations and simplify the integration process.

Coordinate Transformation

Let's apply the transformation:

( x u ) ( y 4v ) ( z 7w )

Under this transformation, the integral becomes:

( iiint_{Omega} x , dx , dy , dz J_{xyz} , int int int u , du , dv , dw )

where ( J_{xyz} ) is the Jacobian determinant.

Computing the Jacobian Determinant

The Jacobian determinant (J_{xyz}) is given by:

( J_{xyz} left| begin{array}{ccc} frac{partial x}{partial u} frac{partial x}{partial v} frac{partial x}{partial w} frac{partial y}{partial u} frac{partial y}{partial v} frac{partial y}{partial w} frac{partial z}{partial u} frac{partial z}{partial v} frac{partial z}{partial w} end{array} right| 28 )

Therefore, the integral simplifies to:

( iiint_{Omega} x , dx , dy , dz 28 int int int u , du , dv , dw )

Transforming the Region ( Omega )

The region ( Omega ) is described in the original coordinates, but we need to express it in terms of the new coordinates ( u, v, w ).

In the new coordinates, the boundaries are:

( 0 leq v leq 1 ) ( 0 leq u leq v^2 ) ( 0 leq w leq 1 - v )

Evaluating the Integral

We can now evaluate the integral step-by-step:

( 28 int_0^1 int_0^{v^2} int_0^{1-v} u , dw , du , dv )

Since the integrand does not depend on ( w ), we can factor it out:

( 28 int_0^1 int_0^{v^2} u (int_0^{1-v} 1 , dw) , du , dv 28 int_0^1 int_0^{v^2} u (1 - v) , du , dv )

First, integrate with respect to ( u ):

( 28 int_0^1 u (1 - v) left[ frac{u^2}{2} right]_0^{v^2} , dv 28 int_0^1 u (1 - v) frac{(v^2)^2}{2} , dv 28 int_0^1 u (1 - v) frac{v^4}{2} , dv )

Let ( u v^2 ), then ( du 2v , dv ):

( 28 int_0^1 v^2 (1 - v) frac{v^4}{2} , dv 14 int_0^1 v^6 (1 - v) , dv )

Now, integrate with respect to ( v ):

( 14 int_0^1 (v^6 - v^7) , dv 14 left[ frac{v^7}{7} - frac{v^8}{8} right]_0^1 14 left( frac{1}{7} - frac{1}{8} right) 14 left( frac{8 - 7}{56} right) 14 frac{1}{56} frac{1}{4} )

Therefore, the value of the integral is:

( iiint_{Omega} x , dx , dy , dz frac{7}{15} )

Conclusion

By using an effective coordinate transformation and carefully computing the Jacobian determinant, we were able to simplify the integration process and obtain the solution to the given triple integral problem. This method can be applied to similar problems involving complex geometries in three-dimensional space.