Questions: Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded by the curves f(x)=4-x^2 and y=0. The volume is cubic units. (Type an exact answer, using π as needed.)

Solution Steps

To find the volume of the solid of revolution formed by rotating the region bounded by the curves $f(x) = 4 - x^2$ and $y = 0$ about the $x$-axis, we can use the disk method. The volume $V$ is given by the integral:

$$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$$

where $a$ and $b$ are the points where the curve intersects the $x$-axis.

1. Determine the points of intersection of $f(x) = 4 - x^2$ and $y = 0$.
2. Set up the integral using the disk method.
3. Evaluate the integral to find the volume.

To find the points of intersection of the curves $f(x) = 4 - x^2$ and $y = 0$, we solve the equation: $$4 - x^2 = 0$$ This gives: $$x^2 = 4$$ $$x = \pm 2$$ Thus, the points of intersection are $x = -2$ and $x = 2$.

Using the disk method, the volume $V$ of the solid of revolution is given by: $$V = \pi \int_{-2}^{2} (4 - x^2)^2 \, dx$$

We evaluate the integral: $$V = \pi \int_{-2}^{2} (4 - x^2)^2 \, dx$$ $$V = \pi \left[ \frac{512}{15} \right]$$ $$V = \frac{512\pi}{15}$$

The numerical value of the volume is: $$V \approx 107.2330 \, \text{cubic units}$$

Final Answer