Math Problem Statement

Evaluate the triple integral.

E 9z dV, where E is bounded by the cylinder y2 + z2 = 9 and the planes x = 0, y = 3x, and z = 0 in the first octant

Solution

To evaluate the triple integral E9zdV\iiint_E 9z \, dV, where EE is bounded by the cylinder y2+z2=9y^2 + z^2 = 9 and the planes x=0x = 0, y=3xy = 3x, and z=0z = 0 in the first octant, we can follow these steps:

Step 1: Set up the bounds of the integral

  1. Cylinder: The equation y2+z2=9y^2 + z^2 = 9 describes a cylinder with radius 3, extending infinitely along the xx-axis.
  2. Planes: The plane x=0x = 0 bounds EE on the left, y=3xy = 3x bounds EE from above, and z=0z = 0 bounds it from below.
  3. First Octant: We restrict to x,y,z0x, y, z \geq 0.

Step 2: Determine the bounds for xx, yy, and zz

  1. For xx: Since y=3xy = 3x, when yy reaches its maximum value of 3 (the radius of the cylinder), we have: 3=3xx=1.3 = 3x \Rightarrow x = 1. Thus, 0x10 \leq x \leq 1.

  2. For yy: From the plane y=3xy = 3x and the cylinder's projection in the yzyz-plane: 0y3.0 \leq y \leq 3.

  3. For zz: From the cylinder's equation: 0z9y2.0 \leq z \leq \sqrt{9 - y^2}.

Step 3: Set up the triple integral

The triple integral can be expressed as: 0103x09y29zdzdydx.\int_0^1 \int_0^{3x} \int_0^{\sqrt{9 - y^2}} 9z \, dz \, dy \, dx.

Step 4: Evaluate the integral

  1. Integrate with respect to zz: 09y29zdz=9[z22]09y2=9(9y2)2=9(9y2)2.\int_0^{\sqrt{9 - y^2}} 9z \, dz = 9 \left[ \frac{z^2}{2} \right]_0^{\sqrt{9 - y^2}} = 9 \cdot \frac{(9 - y^2)}{2} = \frac{9(9 - y^2)}{2}.

  2. Substituting into the next integral: 0103x9(9y2)2dydx.\int_0^1 \int_0^{3x} \frac{9(9 - y^2)}{2} \, dy \, dx.

  3. Integrate with respect to yy: [ \int_0^{3x} \frac{9(9 - y^2)}{2} , dy = \frac{9}{2} \left[ 9y - \frac{y^3}{3} \right]_0^{3x} = \frac{9}{2} \left[ 27x - \frac{(3x)^3}{3} \right] = \frac{9}{2} \left[ 27x - 9x^3 \right] = \frac{9}{2} \cdot 9x(3 - x^2) = \frac{81x(3 - x^2)}{2}. ]

  4. Now integrate with respect to xx: [ \int_0^1 \frac{81x(3 - x^2)}{2} , dx = \frac{81}{2} \left[ \frac{3x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{81}{2} \left[ \frac{3}{2} - \frac{1}{4} \right] = \frac{81}{2} \left[ \frac{6}{4} - \frac{1}{4} \right] = \frac{81}{2} \cdot \frac{5}{4} = \frac{405}{8}. ]

Final Result

The value of the triple integral is: 4058.\frac{405}{8}.

If you have any questions or need further details, feel free to ask! Here are some related questions to consider:

  1. What other coordinate systems might simplify this integral?
  2. How would the bounds change if the cylinder had a different radius?
  3. Can you explain how to visualize the region EE?
  4. What happens to the integral if we change the integrand to a different function?
  5. How would the result differ if calculated in cylindrical coordinates?

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Math Problem Analysis

Mathematical Concepts

Triple Integrals
Cylindrical Coordinates
Volume of Solids

Formulas

Triple Integral Formula
Volume under Surface

Theorems

Fundamental Theorem of Calculus
Integration Bounds Theorem

Suitable Grade Level

Grades 11-12