Math Problem Statement

Use cylindrical coordinates to calculate \( \iiint_W f(x, y, z) \, dV \) for the given function and region: \( f(x, y, z) = z \), \( x^2 + y^2 \leq 25 \), \( z \leq 25 \)

Solution

To evaluate the integral using cylindrical coordinates, let's first analyze the problem.

Given:

  • The function f(x,y,z)=zf(x, y, z) = z
  • The region of integration is defined by: x2+y225,z25x^2 + y^2 \leq 25, \quad z \leq 25

Step 1: Setting up cylindrical coordinates

In cylindrical coordinates:

  • x=rcosθx = r \cos \theta
  • y=rsinθy = r \sin \theta
  • z=zz = z
  • The volume element dV=rdzdrdθdV = r \, dz \, dr \, d\theta

The region x2+y225x^2 + y^2 \leq 25 describes a circular base of radius 5 in the xyxy-plane. Additionally, z25z \leq 25 defines the height of the region.

Step 2: Setting up the bounds

  1. For rr: 0r50 \leq r \leq 5 (radius of the circular base).
  2. For θ\theta: 0θ2π0 \leq \theta \leq 2\pi (full rotation around the zz-axis).
  3. For zz: 0z250 \leq z \leq 25 (bounded vertically).

Step 3: Write the integral in cylindrical coordinates

The integral becomes: Wf(x,y,z)dV=Wzrdzdrdθ\iiint_W f(x, y, z) \, dV = \iiint_W z \, r \, dz \, dr \, d\theta

Expanding with the limits: 02π05025zrdzdrdθ\int_0^{2\pi} \int_0^5 \int_0^{25} z \, r \, dz \, dr \, d\theta

Step 4: Evaluate the integral

We proceed by integrating step-by-step.

  1. Integrate with respect to zz: 02π05025zrdzdrdθ=02π05[z22]025rdrdθ\int_0^{2\pi} \int_0^5 \int_0^{25} z \, r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^5 \left[ \frac{z^2}{2} \right]_0^{25} r \, dr \, d\theta Evaluating the inner integral: =02π052522rdrdθ= \int_0^{2\pi} \int_0^5 \frac{25^2}{2} \, r \, dr \, d\theta =02π056252rdrdθ= \int_0^{2\pi} \int_0^5 \frac{625}{2} \, r \, dr \, d\theta

  2. Integrate with respect to rr: =02π[6252r22]05dθ= \int_0^{2\pi} \left[ \frac{625}{2} \frac{r^2}{2} \right]_0^5 \, d\theta =02π6252252dθ= \int_0^{2\pi} \frac{625}{2} \cdot \frac{25}{2} \, d\theta =02π156254dθ= \int_0^{2\pi} \frac{15625}{4} \, d\theta

  3. Integrate with respect to θ\theta: =1562542π= \frac{15625}{4} \cdot 2\pi =156252π4= \frac{15625 \cdot 2 \cdot \pi}{4} =7812.5π= 7812.5 \pi

Final Answer:

Wf(x,y,z)dV=7812.5π\iiint_W f(x, y, z) \, dV = 7812.5 \pi

Would you like further details or have any questions?


Related Questions:

  1. How would the setup change if the region was a cone rather than a cylinder?
  2. What is the physical interpretation of this integral if zz represents density?
  3. How do cylindrical coordinates simplify this problem compared to Cartesian coordinates?
  4. How would this integral be evaluated if f(x,y,z)=x+y+zf(x, y, z) = x + y + z?
  5. What are some common applications of triple integrals in cylindrical coordinates?

Tip:

When solving problems in cylindrical coordinates, always carefully set up the bounds for rr, θ\theta, and zz based on the geometry of the region.

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Math Problem Analysis

Mathematical Concepts

Cylindrical Coordinates
Triple Integral
Volume Integration

Formulas

Cylindrical coordinate conversion: \( x = r \cos \theta \), \( y = r \sin \theta \), \( z = z \)
Volume element in cylindrical coordinates: \( dV = r \, dz \, dr \, d\theta \)

Theorems

Fundamental Theorem of Calculus for Triple Integrals

Suitable Grade Level

Undergraduate (Calculus III or Multivariable Calculus)