We need to find the area of the region bounded by the curve \(y^2 = 8x\) and the line \(y = 2x\).
First, find the points of intersection by setting the equations equal to each other:
\(y^2 = 8x\) and \(y = 2x\). Substitute \(y = 2x\) into the first equation: \((2x)^2 = 8x\)
\(4x^2 = 8x\)
\(4x^2 - 8x = 0\)
\(4x(x - 2) = 0\)
\(x = 0\) or \(x = 2\)
When \(x = 0\), \(y = 2(0) = 0\).
When \(x = 2\), \(y = 2(2) = 4\).
So, the points of intersection are (0, 0) and (2, 4).
Now we set up the integral for the area between the curves. Since \(y^2 = 8x\), \(x = \frac{y^2}{8}\).
We want to integrate with respect to y, so we express x in terms of y. The line is \(x = \frac{y}{2}\).
We integrate from y = 0 to y = 4. The area is given by:
\(\int_{0}^{4} (\frac{y}{2} - \frac{y^2}{8}) \, dy\)
\(= [\frac{y^2}{4} - \frac{y^3}{24}]_{0}^{4}\)
\(= (\frac{4^2}{4} - \frac{4^3}{24}) - (0 - 0)\)
\(= (\frac{16}{4} - \frac{64}{24})\)
\(= 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}\)
Therefore, the area of the region is \(\frac{4}{3}\) square units.
Thus, the correct option is (B) \(\frac{4}{3}\) sq. units.
We want to find the area of the region bounded by the curve $ y^2 = 8x $ and the line $ y = 2x $.
First, find the intersection points:
$$ (2x)^2 = 8x \implies 4x^2 = 8x \implies 4x^2 - 8x = 0 \implies 4x(x-2) = 0 \implies x=0 \text{ or } x=2. $$
When $ x=0 $, $ y=2(0) = 0 $. When $ x=2 $, $ y=2(2) = 4 $. The intersection points are $ (0,0) $ and $ (2,4) $.
The area is given by:
$$ \int_0^2 (\sqrt{8x} - 2x) dx. $$
Split the integral:
$$ \int_0^2 (\sqrt{8x} - 2x) dx = \sqrt{8} \int_0^2 x^{1/2} dx - 2 \int_0^2 x dx. $$
Evaluate each term separately. For the first term:
$$ \sqrt{8} \int_0^2 x^{1/2} dx = 2\sqrt{2} \cdot \frac{x^{3/2}}{3/2} \Big|_0^2 = 2\sqrt{2} \cdot \frac{2}{3} \cdot (2^{3/2} - 0) = 2\sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2} = \frac{16}{3}. $$
For the second term:
$$ 2 \int_0^2 x dx = 2 \cdot \frac{x^2}{2} \Big|_0^2 = (2^2 - 0) = 4. $$
Combine the results:
$$ \int_0^2 (\sqrt{8x} - 2x) dx = \frac{16}{3} - 4 = \frac{16}{3} - \frac{12}{3} = \frac{4}{3}. $$
The area of the region bounded by the curve and the line is $ \frac{4}{3} $ square units.
The eccentricity of the curve represented by $ x = 3 (\cos t + \sin t) $, $ y = 4 (\cos t - \sin t) $ is: