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Math  /  Geometry

QuestionFind the area of the following region. The region common to the circles r=−10cos⁡θr=-10 \cos \theta and r=5r=5.

Studdy Solution

STEP 1

1. The problem involves polar coordinates.
2. The equation r=−10cos⁡θ r = -10 \cos \theta represents a circle.
3. The equation r=5 r = 5 represents a circle.
4. We are finding the area of the region common to both circles.

STEP 2

1. Identify the equations of the circles.
2. Determine the points of intersection.
3. Set up the integral for the area of the common region.
4. Evaluate the integral to find the area.

STEP 3

Identify the equations of the circles:
- The circle r=−10cos⁡θ r = -10 \cos \theta can be rewritten in Cartesian coordinates as (x+5)2+y2=25 (x + 5)^2 + y^2 = 25 . It is a circle centered at (−5,0) (-5, 0) with radius 5 5 . - The circle r=5 r = 5 is centered at the origin with radius 5 5 .

STEP 4

Determine the points of intersection:
Set the equations equal to find the points of intersection:
−10cos⁡θ=5 -10 \cos \theta = 5
Solve for θ \theta :
cos⁡θ=−12 \cos \theta = -\frac{1}{2}
The solutions for θ \theta are:
θ=2π3,4π3 \theta = \frac{2\pi}{3}, \frac{4\pi}{3}

STEP 5

Set up the integral for the area of the common region:
The area A A of the region common to both circles can be found by integrating the difference of the areas in polar coordinates from θ=2π3 \theta = \frac{2\pi}{3} to θ=4π3 \theta = \frac{4\pi}{3} :
A=12∫2π34π3(52−(−10cos⁡θ)2) dθ A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left(5^2 - (-10 \cos \theta)^2 \right) \, d\theta

STEP 6

Evaluate the integral to find the area:
Calculate the integral:
A=12∫2π34π3(25−100cos⁡2θ) dθ A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left(25 - 100 \cos^2 \theta \right) \, d\theta
Use the identity cos⁡2θ=1+cos⁡2θ2 \cos^2 \theta = \frac{1 + \cos 2\theta}{2} to simplify:
A=12∫2π34π3(25−50(1+cos⁡2θ)) dθ A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left(25 - 50(1 + \cos 2\theta) \right) \, d\theta
=12∫2π34π3(−25−50cos⁡2θ) dθ = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left(-25 - 50 \cos 2\theta \right) \, d\theta
=12[−25θ−25sin⁡2θ]2π34π3 = \frac{1}{2} \left[ -25\theta - 25 \sin 2\theta \right]_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}
Evaluate the definite integral:
A=12[−25(4π3−2π3)−25(sin⁡8π3−sin⁡4π3)] A = \frac{1}{2} \left[ -25\left(\frac{4\pi}{3} - \frac{2\pi}{3}\right) - 25 (\sin \frac{8\pi}{3} - \sin \frac{4\pi}{3}) \right]
=12[−25(2π3)] = \frac{1}{2} \left[ -25\left(\frac{2\pi}{3}\right) \right]
=12[−50π3] = \frac{1}{2} \left[ -\frac{50\pi}{3} \right]
=−25π3 = -\frac{25\pi}{3}
Since the area must be positive, we take the absolute value:
A=25π3 A = \frac{25\pi}{3}
The area of the region common to both circles is:
25π3 \boxed{\frac{25\pi}{3}}

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