Math Problem Statement

isn't it 0 to 2pi

Solution

You are correct that the angle θ\theta should range from 00 to 2π2\pi, as the curve r=5cosθr = 5\cos\theta traces a complete cardioid over this interval. Let's correct the computation with this range.

Corrected Approach:

Step 1: Set up the integral

The volume integral in cylindrical coordinates remains the same: V=θ=02πr=05cosθz=02rsinθrdzdrdθV = \int_{\theta=0}^{2\pi} \int_{r=0}^{5\cos\theta} \int_{z=0}^{-2r\sin\theta} r \, dz \, dr \, d\theta

Step 2: Integrate with respect to zz:

z=02rsinθrdz=r[2rsinθ]=2r2sinθ\int_{z=0}^{-2r\sin\theta} r \, dz = r[-2r\sin\theta] = -2r^2\sin\theta

Step 3: Integrate with respect to rr:

r=05cosθ2r2sinθdr=2sinθ05cosθr2dr=2sinθ(5cosθ)33=250cos3θsinθ3\int_{r=0}^{5\cos\theta} -2r^2\sin\theta \, dr = -2\sin\theta \int_{0}^{5\cos\theta} r^2 \, dr = -2\sin\theta \cdot \frac{(5\cos\theta)^3}{3} = -\frac{250\cos^3\theta\sin\theta}{3}

Step 4: Integrate with respect to θ\theta:

Now, we integrate from θ=0\theta = 0 to 2π2\pi: V=02π250cos3θsinθ3dθV = \int_{0}^{2\pi} -\frac{250\cos^3\theta\sin\theta}{3} \, d\theta

Step 5: Solve the integral

Notice the integrand 250cos3θsinθ-250\cos^3\theta\sin\theta is an odd function of sinθ\sin\theta (since it’s multiplied by cos3θ\cos^3\theta). When integrated over a full period [0,2π][0, 2\pi], the positive and negative contributions cancel out.

Thus, the volume is: V=0V = 0

Conclusion:

The corrected volume of the solid is indeed 0\boxed{0}, which occurs due to the symmetry of the function over the interval [0,2π][0, 2\pi]. The positive and negative contributions to the volume cancel each other out.

Do you have any more questions, or would you like to dive into a different aspect of the problem?

Related Questions:

  1. What role does symmetry play in definite integrals?
  2. How do you recognize odd and even functions in integrals?
  3. What is the significance of the bounds in polar coordinates?
  4. How does changing bounds affect the outcome of integrals?
  5. What are common cases where volume integrals yield zero?

Tip:

Always consider the symmetry and periodicity of the integrand when computing volumes or areas, especially in problems involving trigonometric functions.

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Math Problem Analysis

Mathematical Concepts

Polar coordinates
Volume integrals
Symmetry in integrals

Formulas

Volume integral in cylindrical coordinates

Theorems

Cancellation of odd functions over symmetric intervals

Suitable Grade Level

Advanced Undergraduate