Question Statement

Find the volume of the solid that is obtained when the region under the curve $y=\sqrt{x}$ over the interval $[1,4]$ is revolved about the $x$-axis.

---

Background and Explanation

This problem involves calculating a volume of revolution using the disk method. When a region bounded by a curve $y=f(x)$ is rotated about the $x$-axis, it generates a solid whose volume can be found by integrating the area of circular cross-sections perpendicular to the axis of rotation.

---

Solution

To find the volume, we apply the disk method formula for rotation about the $x$-axis. The radius of each disk at position $x$ is given by the function value $f(x)=\sqrt{x}$.

First, set up the volume integral using the disk method formula:

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

Substituting the given function $f(x)=\sqrt{x}$ and the interval $[1,4]$:

$V=\pi {\int }_1^4(\sqrt{x}{)}^{2}dx$

Simplify the integrand by squaring the function. Since $(\sqrt{x}{)}^2=x$:

$V=\pi {\int }_1^{4}xdx$

Find the antiderivative of $x$ using the power rule for integration:

$V=\pi {\left[\frac{x^2}{2}\right]}_1^4$

Evaluate the definite integral by substituting the upper and lower bounds:

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

Calculate the values:

$V=\frac{\pi }{2}(16-1)$

Simplify to obtain the final volume:

$V=\frac{15\pi }{2}$

---

Key Formulas or Methods Used

• Disk Method (rotation about x-axis): $V=\pi {\int }_a^b[f(x){]}^{2}dx$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is an antiderivative of $f$

---

Summary of Steps

1. Identify the method: Recognize this as a disk method problem for volume of revolution about the $x$-axis.
2. Set up the integral: Write $V=\pi {\int }_1^4(\sqrt{x}{)}^{2}dx$.
3. Simplify the integrand: Reduce $(\sqrt{x}{)}^2$ to $x$.
4. Integrate: Find the antiderivative $\frac{x^2}{2}$.
5. Evaluate bounds: Substitute limits to get $\frac{\pi }{2}(16-1)$.
6. Calculate: Simplify to the final answer $\frac{15\pi }{2}$ cubic units.