The parabolic arc y = √x, 1 ≤ x ≤ 2 is revolved around the x-axis. The volume of the solid of revolution is

Volume of Revolution Explained

When a two-dimensional region is revolved around an axis, it forms a three-dimensional solid. The volume of such a solid, known as a solid of revolution, can be calculated using integration. For revolution around the x-axis, the most common method is the Disk Method.

The parabolic arc given is $y=\sqrt{x}$, and it is revolved around the x-axis over the interval $1\le x\le 2$.

Formula for Volume of Revolution

The volume $V$ of a solid generated by revolving the region under the curve $y=f(x)$ from $x=a$ to $x=b$ around the x-axis is given by the formula:

$$V=\pi {\int }_a^b[f(x){]}^{2}dx$$

Applying the Volume Formula to the Parabolic Arc

In this specific problem:

- The function is $f(x)=\sqrt{x}$.
- The square of the function is $[f(x){]}^2=(\sqrt{x}{)}^2=x$.
- The lower limit of integration (a) is $1$.
- The upper limit of integration (b) is $2$.

Substitute these values into the volume formula:

$$V=\pi {\int }_1^{2}xdx$$

Calculating the Definite Integral

Now, we need to evaluate the definite integral.

1. **Integrate $x$:** The integral of $x$ with respect to $x$ is $\frac{x^2}{2}$.
2. Apply the Limits of Integration: Evaluate the antiderivative at the upper limit (2) and subtract its value at the lower limit (1).

$$V=\pi {\left[\frac{x^2}{2}\right]}_1^2$$

$$V=\pi \left(\frac{(2{)}^2}{2}-\frac{(1{)}^2}{2}\right)$$

$$V=\pi \left(\frac{4}{2}-\frac{1}{2}\right)$$

$$V=\pi \left(2-\frac{1}{2}\right)$$

To subtract the fractions, find a common denominator:

$$V=\pi \left(\frac{4}{2}-\frac{1}{2}\right)$$

$$V=\pi \left(\frac{4-1}{2}\right)$$

$$V=\pi \left(\frac{3}{2}\right)$$

$$V=\frac{3\pi }{2}$$

Final Volume of the Solid

The volume of the solid of revolution generated by revolving the parabolic arc $y=\sqrt{x}$ from $x=1$ to $x=2$ around the x-axis is $\frac{3\pi }{2}$ cubic units.