Question

Find the area of the region which lies outside of the curve r=10+10sintheta and inside of the curve r=30sintheta

Answer 1

To find the area of the region which lies outside of the curve r=10+10sintheta and inside of the curve r=30sintheta, determine the points of intersection between the two curves, find the equation of the curves in Cartesian coordinates, and integrate the difference between the y-values with respect to x to find the area.

Step-by-step explanation:

To find the area of the region which lies outside of the curve r=10+10sintheta and inside of the curve r=30sintheta, we need to determine the points of intersection between the two curves. To do this, we can set the equations equal to each other: 10+10sintheta = 30sintheta. Solving for theta, we get theta = pi/6 and theta = 7pi/6. Next, we can integrate the difference between the two curves from theta = pi/6 to theta = 7pi/6 to find the area.

First, let's find the equation of the curves in Cartesian coordinates. r = 10 + 10sin(theta) can be expressed as x = (10 + 10sin(theta))cos(theta) and y = (10 + 10sin(theta))sin(theta). Similarly, r = 30sin(theta) can be expressed as x = (30sin(theta))cos(theta) and y = (30sin(theta))sin(theta). By substituting theta = pi/6 and theta = 7pi/6 into the equations, we can find the corresponding (x, y) coordinates.

Next, we can calculate the area between the curves by integrating the difference between the two y-values with respect to x from the x-coordinate of the smaller curve to the larger curve. The area can be given by the integral of (y1 - y2) dx, where y1 is the y-coordinate of the larger curve and y2 is the y-coordinate of the smaller curve, and the limits of integration are the x-coordinates of the points of intersection. By integrating the equation with these limits, we can determine the area of the region.

Answer 2

The area of the region is 75π square units.

Visualize the curves:

r = 10 + 10 sin θ represents a limaçon with a loop.

r = 30 sin θ represents a cardioid.

Find the intersection points:

Set the two equations equal to each other: 10 + 10 sin θ = 30 sin θ

Simplify: sin θ = 1/2

The solutions are θ = π/6 and θ = 5π/6.

Set up the integral:

We'll use the formula for the area enclosed by a polar curve:

A = (1/2) ∫[a, b] (r₂(θ)² - r₁(θ)²) dθ

r₂(θ) = 30 sin θ (the outer curve)

r₁(θ) = 10 + 10 sin θ (the inner curve)

a = π/6 (the first intersection point)

b = 5π/6 (the second intersection point)

Evaluate the integral:

A = (1/2) ∫[π/6, 5π/6] (30 sin θ)² - (10 + 10 sin θ)² dθ

= (1/2) ∫[π/6, 5π/6] (900 sin² θ - 100 - 200 sin θ - 100 sin² θ) dθ

= (1/2) ∫[π/6, 5π/6] (800 sin² θ - 200 sin θ - 100) dθ

= 75π