Math Problem Statement

Question (4): Evaluate the following Integrals: 1) ∭ y 𝑑𝑉 𝑄 , where 𝑄 is the β€œTetrahedron” bounded by the planes: π‘₯ = 0, 𝑦 = 0, 𝑧 = 0 and π‘₯ + 𝑦 + 𝑧 = 1

Solution

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

Steps to Solve:

  1. Determine the Region QQ:

    • The tetrahedron is defined by the planes x=0x = 0, y=0y = 0, z=0z = 0, and x+y+z=1x + y + z = 1.
    • The bounds for xx, yy, and zz can be found from these planes.
  2. Set Up the Integral:

    • The plane x+y+z=1x + y + z = 1 implies that for fixed xx and yy, zz ranges from 00 to 1βˆ’xβˆ’y1 - x - y.
    • For fixed xx, yy ranges from 00 to 1βˆ’x1 - x.
    • xx ranges from 00 to 11.

Thus, the integral can be written as: ∭Qy dV=∫01∫01βˆ’x∫01βˆ’xβˆ’yy dz dy dx\iiint_Q y \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} y \, dz \, dy \, dx

  1. Evaluate the Innermost Integral (with respect to zz): ∫01βˆ’xβˆ’yy dz=y∫01βˆ’xβˆ’ydz=y[z]01βˆ’xβˆ’y=y(1βˆ’xβˆ’y)\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)

  2. Evaluate the Second Integral (with respect to yy): ∫01βˆ’xy(1βˆ’xβˆ’y) dy=∫01βˆ’x(yβˆ’xyβˆ’y2) dy\int_{0}^{1-x} y(1-x-y) \, dy = \int_{0}^{1-x} (y - xy - y^2) \, dy Evaluate each term separately: ∫01βˆ’xy dy=[y22]01βˆ’x=(1βˆ’x)22\int_{0}^{1-x} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1-x} = \frac{(1-x)^2}{2} ∫01βˆ’xxy dy=x∫01βˆ’xy dy=x[y22]01βˆ’x=xβ‹…(1βˆ’x)22\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} ∫01βˆ’xy2 dy=[y33]01βˆ’x=(1βˆ’x)33\int_{0}^{1-x} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1-x} = \frac{(1-x)^3}{3} Combine these results: ∫01βˆ’xy(1βˆ’xβˆ’y) dy=(1βˆ’x)22βˆ’xβ‹…(1βˆ’x)22βˆ’(1βˆ’x)33\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} =(1βˆ’x)22(1βˆ’x)βˆ’(1βˆ’x)33=(1βˆ’x)32βˆ’(1βˆ’x)33=(36βˆ’26)(1βˆ’x)3=16(1βˆ’x)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

  3. Evaluate the Outer Integral (with respect to xx): ∫0116(1βˆ’x)3 dx=16∫01(1βˆ’x)3 dx\int_{0}^{1} \frac{1}{6}(1-x)^3 \, dx = \frac{1}{6} \int_{0}^{1} (1-x)^3 \, dx Make a substitution u=1βˆ’xu = 1 - x, thus du=βˆ’dxdu = -dx: =16∫10u3(βˆ’du)=16∫01u3 du= \frac{1}{6} \int_{1}^{0} u^3 (-du) = \frac{1}{6} \int_{0}^{1} u^3 \, du Evaluate the integral: ∫01u3 du=[u44]01=14\int_{0}^{1} u^3 \, du = \left[ \frac{u^4}{4} \right]_{0}^{1} = \frac{1}{4} Therefore: 16β‹…14=124\frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}

Final Answer:

∭Qy dV=124\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 ∭Qx dV\iiint_Q x \, dV for the same tetrahedron.
  2. Evaluate ∭Qz dV\iiint_Q z \, dV for the same tetrahedron.
  3. Find the volume of the tetrahedron using integration.
  4. Evaluate ∭Q(x+y) dV\iiint_Q (x + y) \, dV for the same tetrahedron.
  5. Evaluate ∭Q(y+z) dV\iiint_Q (y + z) \, dV for the same tetrahedron.
  6. Evaluate ∭Q(x+z) dV\iiint_Q (x + z) \, dV for the same tetrahedron.
  7. Find the centroid of the tetrahedron.
  8. Evaluate ∭Q(x2+y2+z2) dV\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.

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Math Problem Analysis

Mathematical Concepts

Triple Integrals
Tetrahedron
Volume Calculation

Formulas

Triple integral setup
Volume of a tetrahedron

Theorems

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Suitable Grade Level

Advanced Mathematics