Math Problem Statement

Evaluate the double integral ∫∫_R f(r, θ) dA with f(r, θ) = r cos(θ), over the region R defined by r from 0 to 5 and θ from 0 to π.

Solution

To evaluate the given double integral, let's go through the problem step-by-step.

The integral is written in polar coordinates as:

0π05rcos(θ)rdrdθ\int_0^{\pi} \int_0^5 r \cos(\theta) \, r \, dr \, d\theta

The function f(r,θ)=rcos(θ)f(r, \theta) = r \cos(\theta) is multiplied by rr due to the Jacobian determinant when converting to polar coordinates.

Step 1: Simplify the Integrand

The integrand becomes: rrcos(θ)=r2cos(θ)r \cdot r \cos(\theta) = r^2 \cos(\theta)

Thus, the integral can be rewritten as: 0π05r2cos(θ)drdθ\int_0^{\pi} \int_0^5 r^2 \cos(\theta) \, dr \, d\theta

Step 2: Integrate with Respect to rr

Now, we focus on the inner integral with respect to rr: 0π(05r2cos(θ)dr)dθ\int_0^{\pi} \left( \int_0^5 r^2 \cos(\theta) \, dr \right) d\theta

Since cos(θ)\cos(\theta) is independent of rr, we can factor it out: =0πcos(θ)(05r2dr)dθ= \int_0^{\pi} \cos(\theta) \left( \int_0^5 r^2 \, dr \right) d\theta

Evaluate the inner integral: 05r2dr=[r33]05=533=1253\int_0^5 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^5 = \frac{5^3}{3} = \frac{125}{3}

Thus, the integral becomes: =0πcos(θ)1253dθ=12530πcos(θ)dθ= \int_0^{\pi} \cos(\theta) \cdot \frac{125}{3} \, d\theta = \frac{125}{3} \int_0^{\pi} \cos(\theta) \, d\theta

Step 3: Integrate with Respect to θ\theta

Now, evaluate the remaining integral: 0πcos(θ)dθ\int_0^{\pi} \cos(\theta) \, d\theta

The integral of cos(θ)\cos(\theta) is sin(θ)\sin(\theta): 0πcos(θ)dθ=[sin(θ)]0π=sin(π)sin(0)=00=0\int_0^{\pi} \cos(\theta) \, d\theta = \left[ \sin(\theta) \right]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0

Final Answer

Since this last integral evaluates to zero, the value of the entire double integral is: 0π05r2cos(θ)drdθ=0\int_0^{\pi} \int_0^5 r^2 \cos(\theta) \, dr \, d\theta = 0

Region RR

The region RR in polar coordinates is defined by:

  • rr ranging from 0 to 5 (a radius up to 5 units)
  • θ\theta ranging from 0 to π\pi (a half-circle in the first and second quadrants)

So, RR represents the upper half of a disk with radius 5.

Would you like a more detailed breakdown of any part of this solution, or do you have questions about the region sketch?


Related Questions:

  1. What is the significance of using polar coordinates in double integrals?
  2. How would the integral change if the limits for θ\theta were from 00 to 2π2\pi?
  3. What is the purpose of the Jacobian in coordinate transformations?
  4. How do we determine the appropriate region RR in polar coordinates?
  5. Can we convert any double integral from Cartesian to polar coordinates?

Tip: When integrating in polar coordinates, always remember to include rr in the integrand to account for the area element dA=rdrdθdA = r \, dr \, d\theta.

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Math Problem Analysis

Mathematical Concepts

Double Integration
Polar Coordinates
Jacobian Determinant
Area of a Region

Formulas

Double integral in polar coordinates: ∫∫_R f(r, θ) dA = ∫_a^b ∫_c^d f(r, θ) r dr dθ
Integration of r^2 and cos(θ) terms

Theorems

Jacobian determinant for polar coordinates

Suitable Grade Level

Undergraduate Calculus