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Chapter 8: Problem 91

Surface Area Find the area of the surface formed by revolving the graph of \(y=2 \sqrt{x}\) on the interval \([0,9]\) about the \(x\) -axis.

Short Answer

Expert verified
The surface area of the surface of revolution is \(120\pi\).

Step by step solution

01

Understanding the Surface Area Formula

The general formula for the surface area \(A\) obtained by rotating the curve \(y=f(x)\), from \(x=a\) to \(x=b\), about the x-axis is \(A=2\pi \int_{a}^{b}f(x)\sqrt{1+(f'(x))^2 }dx\). In our case, the function \(f(x)=2\sqrt{x}\) and its derivative is \(f'(x)=\frac{1}{\sqrt{x}}\). Substituting \(f(x)\) and \(f'(x)\) in the surface area formula gives \(A=2\pi\int_{a}^{b}(2\sqrt{x})\sqrt{1+(\frac{1}{\sqrt{x}})^2 }dx\)
02

Plug in the Interval

Now replace \(a\) and \(b\) in the integral with the boundaries 0 and 9, respectively, hence we get \(A=2\pi \int_{0}^{9}(2\sqrt{x})\sqrt{1+(\frac{1}{\sqrt{x}})^2 }dx\)
03

Calculate the Integral and Simplify

It is now a matter of computing the definite integral and multiplying by \(2\pi\). After calculating, we get \(A = 2\pi \times 60 = 120\pi\).

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