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Chapter 8: Problem 95

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
Compute the integral and multiply by 2\(\pi\) to obtain the surface area. Be very careful with integrating a function involving square roots.

Step by step solution

01

Identify the function and its derivative

The function is \(f(x)=2\sqrt{x}\). Its derivative \(f'(x)\) can be found by applying the power rule for differentiation. The derivative of \(f(x)=x^n\) is \(f'(x)=nx^(n-1)\). So \(f'(x)=d/dx (2\sqrt{x})= 2*(1/2)*x^(-1/2)=1/(sqrt{x})\).
02

Set up the integral for the surface area

Substitute \(f(x)=2\sqrt{x}\) and \(f'(x)=1/(sqrt{x})\) into the formula for the surface area of a solid of revolution about the x-axis to obtain \(A= 2 \pi \int_0^9 (2\sqrt{x}*\sqrt{1 + (1/(4x)}) dx\).
03

Evaluate the integral

Simplify the integrand in the equation obtained in step 2 and then evaluate the integral by using standard integral techniques.
04

Compute the surface area of the solid

After computing the integral, multiply by 2\(\pi\) to obtain the surface area A of the solid obtained by revolving the graph of the function y=2\sqrt{x} about the x-axis on the interval [0, 9].

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