Question

The volume of the solid of revolution of the loop of the curve \[ y^2 = x^4(x + 2) \] about the x-axis (round off to 2 decimal places) is .............

Hint:

For volumes of solids of revolution, use the disk method where the volume is computed as \( \pi \int_{a}^{b} (f(x))^2 dx \).

Answer 1

Correct Answer: 6.6

Step 1: Equation of the curve.
The given equation of the curve is: \[ y^2 = x^4(x + 2). \] We need to find the volume of the solid of revolution formed by rotating the curve about the x-axis. This is done using the disk method, where the volume of the solid is given by: \[ V = \pi \int_{a}^{b} \left( y(x) \right)^2 dx. \] We are given the curve in terms of \( y^2 \), so the volume integral becomes: \[ V = \pi \int_{a}^{b} y^2 dx = \pi \int_{a}^{b} x^4(x + 2) dx. \]
Step 2: Find the limits of integration.
The loop of the curve exists for values of \( x \) between \( -1 \) and \( 1 \), which is where the curve intersects the x-axis. Therefore, the limits of integration are from \( x = -1 \) to \( x = 1 \).
Step 3: Compute the integral.
We now compute the integral: \[ V = \pi \int_{-1}^{1} x^4(x + 2) dx. \] Expanding the integrand: \[ V = \pi \int_{-1}^{1} (x^5 + 2x^4) dx. \] Now, we integrate: \[ \int x^5 dx = \frac{x^6}{6}, \quad \int 2x^4 dx = \frac{2x^5}{5}. \] Evaluating the integrals at the limits \( x = -1 \) and \( x = 1 \), we get: \[ V = \pi \left( \left[\frac{1^6}{6} + \frac{2(1^5)}{5}\right] - \left[\frac{(-1)^6}{6} + \frac{2(-1)^5}{5}\right] \right). \] Simplifying this: \[ V = \pi \left( \left[\frac{1}{6} + \frac{2}{5}\right] - \left[\frac{1}{6} - \frac{2}{5}\right] \right) = \pi \left( \frac{4}{5} \right). \] Thus, the volume of the solid is: \[ V = \frac{4\pi}{5}. \] The final value is approximately: \[ V \approx 2.51 \, \text{(rounded to 2 decimal places)}. \]