Set up the triple integral that gives the volume of the solid tetrahedron D formed by the vertices (0,0,0), (1,1,0), (0,1,1), and (0,1,1).
Understand the Problem
The question is asking to set up a triple integral that will represent the volume of a tetrahedron in 3-dimensional space defined by specific vertices. To solve this, we need to express the volume using the appropriate limits for the integration based on the coordinates of the vertices provided.
Answer
The triple integral for the volume of the tetrahedron is $$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz \, dy \, dx $$
Answer for screen readers
The triple integral representing the volume of the tetrahedron is given by:
$$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz , dy , dx $$
Steps to Solve
- Identify the vertices of the tetrahedron The given vertices of the tetrahedron are:
- ( A(0,0,0) )
- ( B(1,1,0) )
- ( C(0,1,1) )
- ( D(0,0,1) )
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Determine the projection onto a coordinate plane The tetrahedron lies within the volume bounded by these points. We can consider projecting these points onto the ( xy )-plane to establish limits. The lowest plane is defined by ( z = 0 ) and the highest plane varies based on ( x ) and ( y ).
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Establish limits for integration We analyze the vertices to establish the limits:
- The range for ( z ) is from ( 0 ) to a maximum value determined by the vertices.
- For fixed ( z ), the shape forms a triangle in the ( xy )-plane:
- The projection onto the ( xy )-plane gives a triangle with vertices ( (0,0) ), ( (1,1) ), and ( (0,1) ).
The inequalities for ( x ) and ( y ) given the projection are:
- For ( y ) from ( 0 ) to ( 1 - x ) (determined by the line connecting ( (1,1) ) to ( (0,1) )).
- The range for ( x ) is from ( 0 ) to ( 1 ).
- Set up the integral The triple integral can be set up as follows:
$$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz , dy , dx $$
- Impact of the limits on volume In this setup, ( z ) is integrated from ( 0 ) to ( 1 - x - y ) based on the linear relationship defined by the vertices, which reflects the top face of the tetrahedron.
The triple integral representing the volume of the tetrahedron is given by:
$$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz , dy , dx $$
More Information
This integral captures the volume beneath the plane defined by the vertices of the tetrahedron and above the ( xy )-plane. Evaluating this integral will yield the volume of the tetrahedron.
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