Let's break down the problem step-by-step: ## **Step 1: Understand the Region \( E \)** - The region \( E \) is bounded by the coordinate planes \( x=0 \), \( y=0 \), \( z=0 \) and the plane \( x+y+z=1 \). - So, \( E \) is the tetrahedron with vertices at \( (0,0,0) \), \( (1,0,0) \), \( (0,1,0) \), and \( (0,0,1) \). The inequalities describing \( E \) are: \[ x \geq 0, \quad y \geq 0, \quad z \geq 0, \quad x+y+z \leq 1 \] ## **Step 2: Write the Average Value Formula** The average value of \( T(x,y,z) \) over region \( E \) is: \[ T_{\text{avg}} = \frac{1}{\text{Volume}(E)} \iiint_E T(x,y,z) \, dV \] Given: \[ T(x, y, z) = -7x + 8z + 20 \] ## **Step 3: Find the Volume of \( E \)** The volume of a tetrahedron with vertices at \( (0,0,0) \), \( (1,0,0) \), \( (0,1,0) \), and \( (0,0,1) \) is: \[ \text{Volume}(E) = \frac{1}{6} \times \text{Determinant} \] But for this standard tetrahedron, the volume is: \[ \text{Volume}(E) = \frac{1}{6} \] ## **Step 4: Set Up the Triple Integral** The region can be described as: - \( x \) goes from 0 to 1, - \( y \) goes from 0 to \( 1-x \), - \( z \) goes from 0 to \( 1-x-y \): So, \[ \iiint_E T(x, y, z) \, dV = \int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{1-x-y} (-7x + 8z + 20) \, dz \, dy \, dx \] ## **Step 5: Evaluate the Integral** ### Integrate with respect to \( z \): \[ \int_{z=0}^{1-x-y} (-7x + 8z + 20) dz \] Integrate term-by-term: - \( \int (-7x) dz = -7x z \) - \( \int 8z dz = 4z^2 \) - \( \int 20 dz = 20z \) Plug in limits \( z=0 \) to \( z=1-x-y \): \[ = -7x(1-x-y) + 4(1-x-y)^2 + 20(1-x-y) - [0 + 0 + 0] \] \[ = -7x(1-x-y) + 4(1-x-y)^2 + 20(1-x-y) \] ### Expand and combine: Let’s combine terms: \[ -7x(1-x-y) + 4(1-x-y)^2 + 20(1-x-y) \] Let \( a = 1-x-y \): - \( -7x a + 4a^2 + 20a \) Now integrate with respect to \( y \) and then \( x \): \[ \int_{x=0}^1 \int_{y=0}^{1-x} \left[ -7x(1-x-y) + 4(1-x-y)^2 + 20(1-x-y) \right] dy dx \] Let’s expand the terms before integrating: \[ = \int_{x=0}^1 \int_{y=0}^{1-x} [-7x(1-x-y) + 4(1-x-y)^2 + 20(1-x-y)] dy dx \] Let’s write \( (1-x-y)^2 = (1-x)^2 - 2(1-x)y + y^2 \) if needed, but let's integrate term by term. #### First, split the integral: \[ = \int_{x=0}^1 \int_{y=0}^{1-x} \left[ -7x(1-x-y) \right] dy dx + \int_{x=0}^1 \int_{y=0}^{1-x} \left[ 4(1-x-y)^2 \right] dy dx + \int_{x=0}^1 \int_{y=0}^{1-x} \left[ 20(1-x-y) \right] dy dx \] Let's integrate each: --- ### **First term**: \( -7x(1-x-y) \) \[ \int_{y=0}^{1-x} -7x(1-x-y) dy = -7x \int_{y=0}^{1-x} (1-x-y) dy \] Set \( b = 1-x \): \[ = -7x \int_{y=0}^{b} (b-y) dy = -7x \left[ b y - \frac{1}{2} y^2 \right]_{y=0}^{y=b} \] \[ = -7x \left[ b^2 - \frac{1}{2} b^2 \right] = -7x \left( \frac{1}{2} b^2 \right) = -\frac{7x}{2} (1-x)^2 \] --- ### **Second term**: \( 4(1-x-y)^2 \) \[ \int_{y=0}^{1-x} 4(1-x-y)^2 dy \] Let \( u = 1-x-y, \ du = -dy \rightarrow dy = -du \), limits: when \( y=0, u=1-x \), when \( y=1-x, u=0 \): So, \[ \int_{y=0}^{1-x} 4(1-x-y)^2 dy = \int_{u=1-x}^{0} 4u^2 (-du) = \int_{u=0}^{1-x} 4u^2 du = 4 \int_{u=0}^{1-x} u^2 du = 4 \left[ \frac{u^3}{3} \right]_{0}^{1-x} = \frac{4}{3} (1-x)^3 \] --- ### **Third term**: \( 20(1-x-y) \) \[ \int_{y=0}^{1-x} 20(1-x-y) dy = 20 \int_{y=0}^{1-x} (1-x-y) dy = 20 \left[ (1-x)y - \frac{1}{2} y^2 \right]_{0}^{1-x} \] \[ = 20 \left[ (1-x)(1-x) - \frac{1}{2} (1-x)^2 \right] = 20 \left[ (1-x)^2 - \frac{1}{2} (1-x)^2 \right] = 20 \left( \frac{1}{2} (1-x)^2 \right) = 10(1-x)^2 \] --- ### **Sum the results:** Combine all: \[ I = \int_{x=0}^1 \left[ -\frac{7x}{2}(1-x)^2 + \frac{4}{3}(1-x)^3 + 10(1-x)^2 \right] dx \] --- ## **Step 6: Integrate with respect to \( x \)** Let’s expand: - \( (1-x)^2 = 1 - 2x + x^2 \) - \( (1-x)^3 = 1 - 3x + 3x^2 - x^3 \) ### **First term:** \[ -\frac{7x}{2}(1-x)^2 = -\frac{7x}{2}(1 - 2x + x^2) = -\frac{7x}{2} + 7x^2 - \frac{7x^3}{2} \] ### **Second term:** \[ \frac{4}{3}(1-x)^3 = \frac{4}{3}(1 - 3x + 3x^2 - x^3) = \frac{4}{3} - 4x + 4x^2 - \frac{4}{3}x^3 \] ### **Third term:** \[ 10(1-x)^2 = 10(1-2x+x^2) = 10 - 20x + 10x^2 \] --- Now, sum all terms: \[ -\frac{7x}{2} + 7x^2 - \frac{7x^3}{2} + \frac{4}{3} - 4x + 4x^2 - \frac{4}{3}x^3 + 10 - 20x + 10x^2 \] Group like terms: - Constant: \( \frac{4}{3} + 10 = \frac{4}{3} + \frac{30}{3} = \frac{34}{3} \) - \( x \) terms: \( -\frac{7}{2}x -4x -20x = -\frac{7}{2}x -24x = -\frac{7+48}{2}x = -\frac{55}{2}x \) - \( x^2 \) terms: \( 7x^2 + 4x^2 + 10x^2 = 21x^2 \) - \( x^3 \) terms: \( -\frac{7}{2}x^3 - \frac{4}{3}x^3 = \left(-\frac{21}{6} - \frac{8}{6}\right)x^3 = -\frac{29}{6}x^3 \) So, the integral is: \[ I = \int_{x=0}^1 \left( \frac{34}{3} - \frac{55}{2} x + 21 x^2 - \frac{29}{6} x^3 \right) dx \] --- ## **Step 7: Integrate** - \( \int_0^1 dx = 1 \) - \( \int_0^1 x dx = \frac{1}{2} \) - \( \int_0^1 x^2 dx = \frac{1}{3} \) - \( \int_0^1 x^3 dx = \frac{1}{4} \) Plug in: \[ I = \frac{34}{3} (1) - \frac{55}{2} \left(\frac{1}{2}\right) + 21 \left(\frac{1}{3}\right) - \frac{29}{6} \left(\frac{1}{4}\right) \] \[ = \frac{34}{3} - \frac{55}{4} + 7 - \frac{29}{24} \] Find a common denominator, 24: - \( \frac{34}{3} = \frac{272}{24} \) - \( \frac{55}{4} = \frac{330}{24} \) - \( 7 = \frac{168}{24} \) - \( \frac{29}{24} = \frac{29}{24} \) \[ I = \frac{272}{24} - \frac{330}{24} + \frac{168}{24} - \frac{29}{24} \] \[ = \frac{272 - 330 + 168 - 29}{24} \] \[ = \frac{81}{24} \] --- ## **Step 8: Average Value** Recall the formula: \[ T_{\text{avg}} = \frac{1}{\text{Volume}(E)} I = \frac{1}{1/6} \cdot \frac{81}{24} = 6 \cdot \frac{81}{24} = \frac{486}{24} = 20.25 \] --- ## **Step 9: Final Answer** ### **The average temperature over the solid is:** \[ \boxed{20.250} \] (Rounded to three decimal places as required.)