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Find the surface area of a surface created by rotating the region bounded by 𝑓(𝑥) = 𝑥2 and the x-axis, on [0,1], about the x-axis

Question

Find the surface area of a surface created by rotating the region bounded by

f(x)=x2 f(x) = x^2

and the x-axis, on [0,1][0,1], about the x-axis.

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Solution

The surface area A of a solid of revolution generated by rotating a curve y = f(x) from x = a to x = b about the x-axis is given by the formula:

A = 2π ∫ from a to b [f(x) * sqrt(1 + (f'(x))^2)] dx

Here, f(x) = x^2 and the interval is [0,1].

First, we need to find the derivative of f(x), f'(x).

f'(x) = 2x

Now, we can substitute f(x) and f'(x) into the formula:

A = 2π ∫ from 0 to 1 [x^2 * sqrt(1 + (2x)^2)] dx

This integral can be a bit tricky to solve. You might want to use a substitution method to solve it. Let u = 1 + 4x^2. Then du = 8x dx. The limits of integration change to u = 1 when x = 0 and u = 5 when x = 1.

So, the integral becomes:

A = 2π ∫ from 1 to 5 [(u - 1)/8 * sqrt(u)] * (1/8) du

Simplify this to:

A = π/32 ∫ from 1 to 5 [u^(3/2) - u^(1/2)] du

Now, this integral can be easily solved by using the power rule for integration. The power rule states that ∫x^n dx = (1/(n+1))x^(n+1).

So, the solution to the integral is:

A = π/32 [(2/5)u^(5/2) - (2/3)u^(3/2)] evaluated from 1 to 5

Plug in the limits of integration:

A = π/32 [(2/5)*5^(5/2) - (2/3)*5^(3/2) - (2/5)*1^(5/2) + (2/3)*1^(3/2)]

Simplify this to get the final answer:

A = π/32 [20sqrt(5) - 10sqrt(5) - 2/5 + 2/3]

A = π/32 [10sqrt(5) - 2/5 + 2/3]

This is the surface area of the surface created by rotating the region bounded by f(x) = x^2 and the x-axis, on [0,1], about the x-axis.

This problem has been solved

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