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).
To find the area enclosed by the polar curve given by the equation \(r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)\), we use the formula for the area \(A\) enclosed by a polar curve: \[A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta\]
Substituting the given \(r\), we have: \[A = \frac{1}{2} \int_0^{2\pi} \left(\frac{2}{\sqrt{\pi}}(1 - \sin \theta)\right)^2 \, d\theta\]
This simplifies to: \[A = \frac{1}{2} \left(\frac{4}{\pi}\right) \int_0^{2\pi} (1 - \sin \theta)^2 \, d\theta\]
Expanding the integrand gives: \[(1 - \sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta\]
Using the identity \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\), the integrand becomes: \[1 - 2\sin \theta + \frac{1 - \cos 2\theta}{2}\]
Integrating term by term, the area calculation becomes: \[A = \frac{2}{\pi} \left(\int_0^{2\pi} 1 \, d\theta - 2\int_0^{2\pi} \sin \theta \, d\theta + \frac{1}{2}\int_0^{2\pi} (1 - \cos 2\theta) \, d\theta\right)\]
Each integral evaluates as follows: \[\int_0^{2\pi} 1 \, d\theta = 2\pi,\] \[\int_0^{2\pi} \sin \theta \, d\theta = 0,\] \[\int_0^{2\pi} 1 \, d\theta = 2\pi,\] \[\int_0^{2\pi} \cos 2\theta \, d\theta = 0\]
Thus, the area \(A\) becomes: \[A = \frac{2}{\pi}\left(2\pi - 0 + \pi\right) = \frac{2}{\pi}(3\pi) = 6\]
The computed area is 6, which falls within the expected range (6,6).
Important Questions from Application Of Definite Integral (Area)
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.
- Let $S_1$ be the plane figure consisting of the points $(x, y)$ given by the inequalities $|x-1| \le 2$ and$|y + 2| \le 3$. Let $S_2$ be the plane figure given by the inequalities $x - y \ge -2$, $y \ge 1$, and $x \le 3$.Let $S$ be the union of $S_1$ and $S_2$. The area of $S$ is
