Final answer:
To find the area enclosed by the curve r = 6(1+cos(theta)), we can use the formula for the area of a polar region. By determining the limits of integration and performing the integral, we find that the area is 9pi - 9 square units.
Step-by-step explanation:
The curve given by the equation r = 6(1+cos(theta)) represents a cardioid shape.
To find the area enclosed by this curve, we can use the formula for the area of a polar region: A = (1/2)∫[a,b] r^2 d(theta). In this case, the limits of integration 'a' and 'b' can be determined by finding the values of theta for which r=0. We know that cos(theta) = -1 when theta = pi, so the limits of integration are pi/2 to 3pi/2.
Let's calculate the area step-by-step:
- Find the square of r: r^2 = 36(1+cos(theta))^2
- Simplify the expression: r^2 = 36(1+2cos(theta)+cos^2(theta))
- Expand the expression using the identity cos^2(theta) = (1+cos(2theta))/2: r^2 = 36(1+2cos(theta)+(1+cos(2theta))/2)
- Simplify further: r^2 = 18 + 36cos(theta) + 18cos(2theta)
- Integrate the expression using the given limits: A = (1/2)∫[pi/2, 3pi/2] (18 + 36cos(theta) + 18cos(2theta)) d(theta)
- Integrate term by term: A = (1/2)(∫[pi/2, 3pi/2] 18 d(theta) + ∫[pi/2, 3pi/2] 36cos(theta) d(theta) + ∫[pi/2, 3pi/2] 18cos(2theta) d(theta))
- Perform the integrals: A = 9pi + 18sin(theta) + 9sin(2theta) evaluated from pi/2 to 3pi/2
- Plug in the limits: A = 9pi + 18sin(3pi/2) + 9sin(3pi) - (9pi + 18sin(pi/2) + 9sin(pi))
- Use the values of sin(theta) and sin(2theta) to simplify further: A = 9pi - 18 + 9
- Final result: A = 9pi - 9 square units