Surface Area When y = √(x + 1) Rotates About the x-Axis

Title: Surface Area When y = √(x + 1) Rotates About the x-Axis Category: mathematics

The curve $y=\sqrt{x+1}$ does not determine one surface area by itself: you also need to know which part of the curve is rotated. The usual interpretation is that the graph is rotated about the $x$-axis over a specified interval $[a,b]$, where $a\geq -1$.

For that interval, the surface area is

$$S=\frac{\pi}{6}\left[(4b+5)^{3/2}-(4a+5)^{3/2}\right].$$

Set up the surface-area integral

When a graph $y=f(x)$ is revolved about the $x$-axis, its surface area over $a\leq x\leq b$ is

$$S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$

The factor $2\pi y$ is the circumference of a circular band at height $y$. The square-root factor accounts for the length of the curve, since a small piece of the graph has length $ds=\sqrt{1+(dy/dx)^2}\,dx$.

For $y=\sqrt{x+1}$, the domain is $x\geq -1$, and

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x+1}}.$$

Substitute both expressions into the formula:

$$\begin{aligned} S&=2\pi\int_a^b \sqrt{x+1}\sqrt{1+\frac{1}{4(x+1)}}\,dx\\ &=2\pi\int_a^b \sqrt{x+1}\sqrt{\frac{4x+5}{4(x+1)}}\,dx\\ &=\pi\int_a^b \sqrt{4x+5}\,dx. \end{aligned}$$

For $x>-1$, both $x+1$ and $4x+5$ are positive, so the simplification is valid there. At $x=-1$ the derivative expression is undefined; the simplified integrand has the finite limit $\pi$, so the integral is interpreted improperly when $a=-1$.

Evaluate the integral

Use $u=4x+5$, so $du=4\,dx$:

$$\int \sqrt{4x+5}\,dx=\frac{1}{6}(4x+5)^{3/2}.$$

Applying the endpoints gives

$$\boxed{S=\frac{\pi}{6}\left[(4b+5)^{3/2}-(4a+5)^{3/2}\right]}.$$

This is the area of the surface generated by the curve segment, not the area of the region under the graph and not the volume of the solid.

Worked example: rotate from x = 0 to x = 3

For the portion of the graph with $0\leq x\leq 3$, take $a=0$ and $b=3$:

$$\begin{aligned} S&=\frac{\pi}{6}\left[(4\cdot3+5)^{3/2}-(4\cdot0+5)^{3/2}\right]\\ &=\frac{\pi}{6}\left[17^{3/2}-5^{3/2}\right]\\ &=\frac{\pi}{6}\left(17\sqrt{17}-5\sqrt{5}\right)\\ &\approx 30.85. \end{aligned}$$

The surface area is about $30.85$ square units. If the intended interval is different, substitute its endpoints for $a$ and $b$ in the boxed formula.

What if the interval starts at x = −1?

The derivative formula contains $1/\sqrt{x+1}$, which is undefined at $x=-1$. This does not make the surface area infinite. The integral can be treated as an improper integral, and its integrand after simplification, $\pi\sqrt{4x+5}$, remains finite there. For example, on $[-1,b]$ the area is

$$S=\frac{\pi}{6}\left[(4b+5)^{3/2}-1\right].$$

You can also avoid the vertical tangent in the graph representation by using $y$ as the parameter. Since $x=y^2-1$, the same curve from $x=-1$ to $x=b$ corresponds to $0\leq y\leq\sqrt{b+1}$. The surface formula becomes $S=2\pi\int y\sqrt{1+(dx/dy)^2}\,dy$, which gives the same result.

Common mistakes

• Leaving out the interval: Without bounds, the problem has no unique finite surface area. The entire curve extends indefinitely, so rotating the whole graph produces an infinite surface area.
• Using the volume formula: The disk method integrates $\pi y^2\,dx$ to find volume. Surface area instead uses $2\pi y\sqrt{1+(y')^2}\,dx$.
• Forgetting the arc-length factor: Integrating just the circumference, $2\pi y$, treats each piece as if the curve were horizontal. The factor $\sqrt{1+(y')^2}$ corrects for the curve’s slope.
• Stopping at the singular derivative: At $x=-1$, handle the integral as improper or parameterize by $y$; the resulting area is finite.