Math Problem Statement

Find the area of ​​the region that is inside the region bounded by the graph of the first equation and outside the region bounded by the graph of the second equation.

r^2=4sen(2theta), r^2=2

Solution

To find the area of the region that is inside the region bounded by the graph of the first equation r2=4sin(2θ)r^2 = 4 \sin(2\theta) and outside the region bounded by the graph of the second equation r2=2r^2 = 2, we need to proceed with the following steps:

Step 1: Interpret the equations

  • The first equation r2=4sin(2θ)r^2 = 4 \sin(2\theta) describes a polar curve, specifically a lemniscate (a figure-eight shape).
  • The second equation r2=2r^2 = 2 describes a circle in polar coordinates with radius 2\sqrt{2}.

Step 2: Determine the bounds of integration

To find the intersection points between the two curves, set the equations equal:

4sin(2θ)=24 \sin(2\theta) = 2

sin(2θ)=12\sin(2\theta) = \frac{1}{2}

The solutions to sin(2θ)=12\sin(2\theta) = \frac{1}{2} within the interval 0θ<2π0 \leq \theta < 2\pi are:

2θ=π6,5π6,13π6,17π62\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}

θ=π12,5π12,13π12,17π12\theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}

However, due to symmetry, we only need to consider one lobe of the lemniscate, i.e., from θ=π12\theta = \frac{\pi}{12} to θ=5π12\theta = \frac{5\pi}{12}.

Step 3: Compute the area

The area of the region between two polar curves r1(θ)r_1(\theta) and r2(θ)r_2(\theta) from θ=a\theta = a to θ=b\theta = b is given by:

A=12ab(r12(θ)r22(θ))dθA = \frac{1}{2} \int_{a}^{b} \left( r_1^2(\theta) - r_2^2(\theta) \right) d\theta

For the region inside r2=4sin(2θ)r^2 = 4 \sin(2\theta) and outside r2=2r^2 = 2, the area becomes:

A=12π125π12(4sin(2θ)2)dθA = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \left( 4 \sin(2\theta) - 2 \right) d\theta

Step 4: Evaluate the integral

To find the exact area, we need to compute the integral:

A=12π125π12(4sin(2θ)2)dθA = \frac{1}{2} \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \left( 4 \sin(2\theta) - 2 \right) d\theta

This can be split into two integrals:

A=2π125π12sin(2θ)dθπ125π122dθA = 2 \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \sin(2\theta) d\theta - \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 2 d\theta

First integral:

sin(2θ)dθ=12cos(2θ)\int \sin(2\theta) d\theta = -\frac{1}{2} \cos(2\theta)

Second integral:

dθ=θ\int d\theta = \theta

So:

[ A = 2 \left[-\frac{1}{2} \cos(2\theta)\right]{\frac{\pi}{12}}^{\frac{5\pi}{12}} - 2 \left[\theta\right]{\frac{\pi}{12}}^{\frac{5\pi}{12}} ]

A=cos(5π6)+cos(π6)2(5π12π12)A = -\cos\left(\frac{5\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right) - 2\left(\frac{5\pi}{12} - \frac{\pi}{12}\right)

A=(32(32))2(4π12)A = \left(\frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{3}}{2}\right)\right) - 2\left(\frac{4\pi}{12}\right)

A=32π3A = \sqrt{3} - \frac{2\pi}{3}

Final Answer:

The area of the region inside the first curve and outside the second curve is 32π3\boxed{\sqrt{3} - \frac{2\pi}{3}}.

Do you want more details or have any questions? Here are some related questions you might consider:

  1. How would the area change if the second curve was r=3r = 3?
  2. What if the first curve was a different polar curve, like r2=4cos(2θ)r^2 = 4 \cos(2\theta)?
  3. Can you calculate the area of a single loop of the lemniscate r2=4sin(2θ)r^2 = 4 \sin(2\theta)?
  4. What are some other applications of polar coordinates in calculating areas?
  5. How does the symmetry of polar curves simplify the area calculation?

Tip: When working with polar coordinates, symmetry can often simplify calculations by reducing the limits of integration.

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Area Between Curves
Trigonometry

Formulas

Area between two polar curves formula

Theorems

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Suitable Grade Level

Undergraduate