Question

The volume of the solid of revolution of the cardioid \(r=a(1+\cos\theta)\) about the initial line is:

• A. \(\pi a^3\)
• B. \(2\pi a^3\)
• C. \(\dfrac{8}{3}\pi a^3\)
• D. \(\dfrac{5}{3}\pi a^3\)

Hint:

For revolving a polar area about the initial line, use \(V=\frac{2\pi}{3}\int r^3\sin\theta\,d\theta\).

Answer 1

The Correct Option is C

Concept:
For a polar curve \(r=f(\theta)\), the volume generated by revolving the area about the initial line is: \[ V=\frac{2\pi}{3}\int_0^\pi r^3\sin\theta\,d\theta \]

Step 1: Write the given curve. \[ r=a(1+\cos\theta) \]

Step 2: Substitute in formula. \[ V=\frac{2\pi}{3}\int_0^\pi a^3(1+\cos\theta)^3\sin\theta\,d\theta \] \[ V=\frac{2\pi a^3}{3}\int_0^\pi (1+\cos\theta)^3\sin\theta\,d\theta \]

Step 3: Put \(u=1+\cos\theta\). \[ du=-\sin\theta\,d\theta \] When: \[ \theta=0,\quad u=2 \] When: \[ \theta=\pi,\quad u=0 \] So, \[ \int_0^\pi (1+\cos\theta)^3\sin\theta\,d\theta =\int_0^2 u^3du \] \[ =\left[\frac{u^4}{4}\right]_0^2 \] \[ =\frac{16}{4}=4 \]

Step 4: Calculate volume. \[ V=\frac{2\pi a^3}{3}\times4 \] \[ V=\frac{8}{3}\pi a^3 \] \[ \therefore \text{Correct Answer is (C)} \]