When calculating the area between curves, always identify the points of intersection and set up the integral based on the difference between the functions. If symmetry exists (as in this case), you can often simplify the calculation by considering only one side and doubling the result. Make sure to properly evaluate the integral and check the units for consistency.
To determine the area enclosed by the curves \(4x^2 = y\) and \(y = 4\), follow these steps:
Rewrite \(4x^2 = y\) as \(x^2 = \frac{y}{4}\), which yields:
\[x = \pm \sqrt{\frac{y}{4}}\]
The intersection point occurs at \(y = 4\). Therefore, calculate the area bounded by these curves from \(y = 0\) to \(y = 4\).
The area is calculated using the following integral:
\[\text{Area} = 2 \int_{0}^{4} \sqrt{\frac{y}{4}} \, dy\]
Simplify the integrand:
\[\text{Area} = 2 \int_{0}^{4} \frac{\sqrt{y}}{2} \, dy = \int_{0}^{4} \sqrt{y} \, dy\]
Evaluate the integral:
\[\int_{0}^{4} y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{2}{3} \times (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3}\]
Consequently, the area enclosed between the curves is \(\frac{16}{3}\) square units.
The eccentricity of the curve represented by $ x = 3 (\cos t + \sin t) $, $ y = 4 (\cos t - \sin t) $ is: