Compute the surface area of revolution about the x-axis over the interval.
y = x^2, [0, 4]
Is it (pi/12)(65^2 - 65)^(3/2)?
Nope. You only integrated 2πx √(1+4x^2)
Each strip of the surface has radius y=x^2, so
∫[0,4] 2πy ds
ds^2 = 1+y'^2 = 1+4x^2, so
A = ∫[0,4] 2π x^2 √(1+4x^2) dx
That's a bit more complicated:
http://www.wolframalpha.com/input/?i=%E2%88%AB[0%2C4]+%282%CF%80+x^2+%E2%88%9A%281%2B4x^2%29%29+dx
To compute the surface area of revolution about the x-axis, we can use the formula:
A = ∫[a,b] 2πf(x)√(1 + (f'(x))^2)dx
In this case, the function is given by f(x) = x^2, and the interval is [0, 4].
First, let's find the derivative of f(x).
f'(x) = 2x
Now, let's substitute these values into the surface area formula:
A = ∫[0,4] 2πx^2√(1 + (2x)^2)dx
Integrating this expression would be a little complex. However, we can use the shell method to simplify it.
According to the shell method, the surface area can be calculated as:
A = 2π ∫[a,b] x * 2πx√(1 + (2x)^2)dx
Now, we can simplify this expression further:
A = 4π^2 ∫[0,4] x^2√(1 + 4x^2)dx
By evaluating this integral, we can find the surface area of revolution about the x-axis over the interval [0, 4]. Letting this expression be S, we have:
S = 4π^2 ∫[0,4] x^2√(1 + 4x^2)dx
Evaluating this integral will give you the correct answer.