Question (4): Evaluate the following Integrals: 1) ∭ y 𝑑𝑉 𝑄 , where 𝑄 is the “Tetrahedron” bounded by the planes: 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 𝑥 + 𝑦 + 𝑧 = 1

To evaluate the integral \(\iiint_Q y \, dV\), where \(Q\) is the tetrahedron bounded by the planes \(x = 0\), \(y = 0\), \(z = 0\), and \(x + y + z = 1\), we need to carefully set up the limits of integration and integrate the function \(y\) over the volume of the tetrahedron.

### Steps to Solve:

1. **Determine the Region \(Q\)**:
   - The tetrahedron is defined by the planes \(x = 0\), \(y = 0\), \(z = 0\), and \(x + y + z = 1\).
   - The bounds for \(x\), \(y\), and \(z\) can be found from these planes.

2. **Set Up the Integral**:
   - The plane \(x + y + z = 1\) implies that for fixed \(x\) and \(y\), \(z\) ranges from \(0\) to \(1 - x - y\).
   - For fixed \(x\), \(y\) ranges from \(0\) to \(1 - x\).
   - \(x\) ranges from \(0\) to \(1\).

Thus, the integral can be written as:
\[
\iiint_Q y \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} y \, dz \, dy \, dx
\]

3. **Evaluate the Innermost Integral** (with respect to \(z\)):
\[
\int_{0}^{1-x-y} y \, dz = y \int_{0}^{1-x-y} dz = y \left[ z \right]_{0}^{1-x-y} = y(1-x-y)
\]

4. **Evaluate the Second Integral** (with respect to \(y\)):
\[
\int_{0}^{1-x} y(1-x-y) \, dy = \int_{0}^{1-x} (y - xy - y^2) \, dy
\]
Evaluate each term separately:
\[
\int_{0}^{1-x} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1-x} = \frac{(1-x)^2}{2}
\]
\[
\int_{0}^{1-x} xy \, dy = x \int_{0}^{1-x} y \, dy = x \left[ \frac{y^2}{2} \right]_{0}^{1-x} = x \cdot \frac{(1-x)^2}{2}
\]
\[
\int_{0}^{1-x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1-x} = \frac{(1-x)^3}{3}
\]
Combine these results:
\[
\int_{0}^{1-x} y(1-x-y) \, dy = \frac{(1-x)^2}{2} - x \cdot \frac{(1-x)^2}{2} - \frac{(1-x)^3}{3}
\]
\[
= \frac{(1-x)^2}{2} (1 - x) - \frac{(1-x)^3}{3} = \frac{(1-x)^3}{2} - \frac{(1-x)^3}{3} = \left( \frac{3}{6} - \frac{2}{6} \right)(1-x)^3 = \frac{1}{6}(1-x)^3
\]

5. **Evaluate the Outer Integral** (with respect to \(x\)):
\[
\int_{0}^{1} \frac{1}{6}(1-x)^3 \, dx = \frac{1}{6} \int_{0}^{1} (1-x)^3 \, dx
\]
Make a substitution \(u = 1 - x\), thus \(du = -dx\):
\[
= \frac{1}{6} \int_{1}^{0} u^3 (-du) = \frac{1}{6} \int_{0}^{1} u^3 \, du
\]
Evaluate the integral:
\[
\int_{0}^{1} u^3 \, du = \left[ \frac{u^4}{4} \right]_{0}^{1} = \frac{1}{4}
\]
Therefore:
\[
\frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}
\]

### Final Answer:
\[
\iiint_Q y \, dV = \frac{1}{24}
\]

Would you like further details or have any questions?

Here are some related questions for further practice:
1. Evaluate \(\iiint_Q x \, dV\) for the same tetrahedron.
2. Evaluate \(\iiint_Q z \, dV\) for the same tetrahedron.
3. Find the volume of the tetrahedron using integration.
4. Evaluate \(\iiint_Q (x + y) \, dV\) for the same tetrahedron.
5. Evaluate \(\iiint_Q (y + z) \, dV\) for the same tetrahedron.
6. Evaluate \(\iiint_Q (x + z) \, dV\) for the same tetrahedron.
7. Find the centroid of the tetrahedron.
8. Evaluate \(\iiint_Q (x^2 + y^2 + z^2) \, dV\) for the same tetrahedron.

**Tip:** When dealing with multiple integrals over a region bounded by planes, carefully determine the limits of integration by analyzing the bounding surfaces.

Learn how to evaluate the triple integral ∭ y dV over a tetrahedron bounded by planes x = 0, y = 0, z = 0, and x + y + z = 1. This problem involves setting up triple integrals, calculating volumes of tetrahedrons, and applying integral calculus techniques. Suitable for advanced mathematics enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation