A curve joining the points $(0,1)$ and $(0,-1)$ is represented parametrically by the equations $x = \sin \theta, y = (1 + \sin \theta)\cos \theta$, where $0\le\theta \le \pi$
Find $\frac{dy}{dx}$ in terms of $\theta$ and determine the x, y coordinates of the points on the curve at which the tangents are parallel to the x-axis and of the point where the tangent is perpendicular to the x-axis. Sketch the curve.
The region in the quadrant $x\ge0, y \ge0$ bounded by the curve and the coordinate axes is rotated about the x-axis through an angle of $2\pi$. Show that the volume swept out is given by $V = \pi \int^1_0 (1+x)^2(1-x^2)dx$. Evaluate V leaving your result in terms of $\pi$.
I calculate $\frac{dy}{dx} = \frac{1-\sin \theta - 2\sin^2\theta}{\cos \theta}$ and also that the required coordinates are:
$(\frac{1}{2}, \pm\frac{3}{4}), (1,0)$
I assume that in order to proceed both sketching the curve and completing the last part of the question I should use the gradient $\frac{dy}{d\theta}$ but cannot find a way to calculate V.
Volume swept by a curve when rotated along x-axis is given as
$$\int_a^b\pi y^2dx$$
Think of it like adding small right cylinders to create the full volume swept
Now $y = (1+\sin\theta)\cos\theta = (1+x)\sqrt{1-x^2}$
$y^2(x) = (1+x)^2(1-x^2)$
Can you proceed?