Find the area of the surface generated by revolving , 0 $\le$ x $\le$ 1 about the x-axis. Approximate area to three decimal places.
A) 16.552
B) 4.138
C) 8.276
D) 11.081
E) 5.327

Question

Find the area of the surface generated by revolving  , 0 $\le$  x $\le$  1 about the x-axis. Approximate area to three decimal places.
A) 16.552 
B) 4.138 
C) 8.276 
D) 11.081 
E) 5.327

Quizplus · Accepted Answer

We can use the formula for the surface area of revolution, which is $\int_{a}^{b}2\pi f(x)\sqrt{1+(f'(x))^2}dx$, where $f(x)$ is the function being revolved, and $a$ and $b$ are the limits of integration.

In this case, $f(x)=x^2$, and we are revolving around the x-axis from $x=0$ to $x=1$. Thus, we need to evaluate the integral $\int_{0}^{1}2\pi x^2\sqrt{1+4x^2}dx$.

Using the substitution $u=1+4x^2$, we can simplify the integral to $\frac{\pi}{4}\int_{1}^{5}\sqrt{u}\,du$, which evaluates to $\frac{\pi}{4}\cdot\frac{2}{3}(5^{3/2}-1^{3/2})\approx 8.276$.

Therefore, the best choice is (C), with an approximate area of 8.276.

Find the Area of the Surface Generated by Revolving ≤\le≤ X ≤\le≤

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