$$ r = 2 - 2 \sin \theta $$
Which one of the following is the area enclosed by the curve for $0 \leq \theta \leq 2\pi$ ?
The problem asks for the area enclosed by the polar curve defined by the equation $ r = 2 - 2 \sin \theta $ over the interval $0 \leq \theta \leq 2\pi$.
Area Calculation Formula
The area $A$ enclosed by a polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ is given by the formula:
$ A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta $In this case, $f(\theta) = 2 - 2 \sin \theta$, $\alpha = 0$, and $\beta = 2\pi$.
Applying the Formula
Substitute the given function and limits into the area formula:
$ A = \frac{1}{2} \int_{0}^{2\pi} (2 - 2 \sin \theta)^2 \, d\theta $Integral Evaluation
- Expand the integrand: $ (2 - 2 \sin \theta)^2 = 4 - 8 \sin \theta + 4 \sin^2 \theta $
- Use trigonometric identity: Replace $\sin^2 \theta$ with $\frac{1 - \cos(2\theta)}{2}$. $ 4 - 8 \sin \theta + 4 \left( \frac{1 - \cos(2\theta)}{2} \right) = 4 - 8 \sin \theta + 2 - 2 \cos(2\theta) = 6 - 8 \sin \theta - 2 \cos(2\theta) $
- Integrate with respect to $\theta$: $ A = \frac{1}{2} \int_{0}^{2\pi} (6 - 8 \sin \theta - 2 \cos(2\theta)) \, d\theta $ $ A = \frac{1}{2} \left[ 6\theta + 8 \cos \theta - \sin(2\theta) \right]_{0}^{2\pi} $
- Evaluate the definite integral: $ A = \frac{1}{2} \left[ \left( 6(2\pi) + 8 \cos(2\pi) - \sin(4\pi) \right) - \left( 6(0) + 8 \cos(0) - \sin(0) \right) \right] $ $ A = \frac{1}{2} \left[ \left( 12\pi + 8(1) - 0 \right) - \left( 0 + 8(1) - 0 \right) \right] $ $ A = \frac{1}{2} [ (12\pi + 8) - 8 ] $ $ A = \frac{1}{2} [ 12\pi ] $ $ A = 6\pi $
Conclusion
The area enclosed by the curve $r = 2 - 2 \sin \theta$ for $0 \leq \theta \leq 2\pi$ is $6\pi$. This corresponds to Option D.
Important Questions from Application Of Definite Integral (Area)
The equation of a closed curve in two-dimensional polar coordinates is given by $r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)$. The area enclosed by the curve is ______ (answer in integer).
The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
Two straight lines pass through the origin $(x_0, y_0) = (0,0)$. One of them passes through the point $(x_1, y_1) = (1,3)$ and the other passes through the point $(x_2, y_2) = (1,2)$.
What is the area enclosed between the straight lines in the interval $[0, 1]$ on the x-axis?
Consider the equation for a curve, $y = f(x) = x^2 + x$.
The area enclosed by the curve, the x -axis ($y = 0$ line); the vertical lines passing through $x = 1$ and $x = 2$ is _________ (rounded off to 2 decimal places)- The area of the region (rounded off to one decimal place) enclosed between the curves $y = x$ and $y = 3\sqrt{x}$ and between the lines $x = 0$ and $x = 1$ is ________ units.
