Math Problem Statement
Solution
To evaluate the triple integral over the solid tetrahedron with vertices , , , and , we first need to determine the limits of integration. These vertices form a tetrahedron in the first octant.
Step 1: Set up the equation of the plane
The vertices are , , , and . We find the equation of the plane containing these points.
For a plane passing through three points , , and , the equation can be given by the determinant:
x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$$ Using the points $$(4, 0, 0)$$, $$(0, 8, 0)$$, and $$(0, 0, 6)$$, the equation of the plane is: $$\begin{vmatrix} x-4 & y & z \\ -4 & 8 & 0 \\ -4 & 0 & 6 \\ \end{vmatrix} = 0$$ Expanding this determinant yields: $$4y - 8z + 6x = 0$$ So, the equation of the plane is $$2x + y + \frac{3}{2}z = 12$$. ### Step 2: Determine the limits of integration Based on this equation, we have: - $$z$$ ranges from $$0$$ to $$6$$. - For a fixed $$z$$, $$y$$ ranges from $$0$$ to $$8 - \frac{4}{3}z$$. - For fixed $$y$$ and $$z$$, $$x$$ ranges from $$0$$ to $$6 - \frac{3}{2}z - \frac{1}{2}y$$. ### Step 3: Setup and evaluate the integral The integral becomes: $$\int_{z=0}^{6} \int_{y=0}^{8-\frac{4}{3}z} \int_{x=0}^{6-\frac{3}{2}z-\frac{1}{2}y} xy \, dx \, dy \, dz$$ Evaluate the innermost integral with respect to $$x$$: $$\int_{0}^{6-\frac{3}{2}z-\frac{1}{2}y} xy \, dx = \left[\frac{1}{2}x^2y\right]_{0}^{6-\frac{3}{2}z-\frac{1}{2}y} = \frac{1}{2}y(6-\frac{3}{2}z-\frac{1}{2}y)^2$$ Evaluate the resulting integral with respect to $$y$$: $$\int_{0}^{8-\frac{4}{3}z} \frac{1}{2}y(6-\frac{3}{2}z-\frac{1}{2}y)^2 \, dy$$ Finally, evaluate the outermost integral with respect to $$z$$: $$\int_{0}^{6} \cdots \, dz$$ ### Conclusion After evaluating these integrals, you will find the value of the triple integral. This involves solving the definite integrals and simplifying to find the final numerical value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Triple Integrals
Geometry of Tetrahedrons
Formulas
Plane equation: ax + by + cz = d
Triple integral: ∫∫∫_E f(x, y, z) dV
Theorems
Determinant method for plane equations
Volume integration in three dimensions
Suitable Grade Level
Undergraduate (Calculus II/III)
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