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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.

Steblo
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  • You may wish to check your differentiation. The equation for $ \ y \ $ is equivalent to $ \ \cos \theta \ + \ \frac12 \sin (2 \theta) \ \ , $ so your numerator in $ \ \frac{1-\sin \theta - 2\sin\theta}{\cos \theta} \ \ $ is not correct, making your result for the first set of coordinates also incorrect. Also, the ratio you have would be $ \ \frac{dy}{dx} \ , $ rather than $ \ \frac{dy}{d\theta} \ \ : $ is the problem statement mistyped? ($ \ \frac{dy}{dx} \ $ is what you would want to work with.) –  Apr 30 '21 at 07:36
  • @boojum Sorry! My typo .I have corrected it now – Steblo Apr 30 '21 at 07:42
  • The $ \ y- $ coordinates for your horizontal tangents need to be fixed, but the trig equation is easy enough to factor... –  Apr 30 '21 at 07:44

1 Answers1

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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?

Dhanvi Sreenivasan
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