Find the area of the surface of revolution generated by revolving the curve y = 3 sqrt (x), 0 <= x <= 4, about the x-axis.
Okay, so I've set up the integral like this:
2pi ∫[0,4] (3 sqrt (x))(sqrt(1+(1/4x)))dx
Which is coming out to 108.5, but that's not giving me the right answer. Can you help me set up the integral correctly or tell me what I'm doing wrong?
okay. You seem to be using shells, so you need to be integrating along y, since the shells have a thickness of dy.
v = ∫[0,6] 2πrh dy
where r = y and h = 4-x = 4-y^2/9
v = ∫[0,6] 2πy(4-y^2/9) dy = 72π
If you want to integrate along x, you need to use discs of thickness dx:
v = ∫[0,4] πr^2 dx
where r=y
v = ∫[0,4] π(3√x)^2 dx = 72π
To find the area of the surface of revolution generated by revolving the curve y = 3√x, 0 ≤ x ≤ 4, about the x-axis, you need to follow a few steps:
1. Determine the element of arc length.
In this case, the element of arc length, ds, is given by:
ds = √(1 + (dy/dx)²) dx
For the given curve, dy/dx = (3/2)√(1/x). Substituting this value into the equation, we get:
ds = √(1 + (3/2)²(1/x)) dx = √(1 + 9/4x) dx
2. Express the element of arc length in terms of x only.
To do this, you need to rewrite ds in terms of x only by substituting for dx using the derivative of x with respect to x (which is 1).
ds = √(1 + 9/4x) dx
3. Set up the integral.
The integral for finding the area of the surface of revolution is:
A = 2π∫[a,b] y ds
In this case, a = 0 and b = 4 since 0 ≤ x ≤ 4.
The curve y = 3√x, so substituting for y and ds, the integral becomes:
A = 2π∫[0,4] (3√x)√(1 + 9/4x) dx
4. Evaluate the integral.
Now you just need to evaluate the integral to find the area.
It seems like you've made a small mistake in setting up the integral. Instead of using sqrt(1 + (1/4x)), you need to use √(1 + 9/4x) in the integrand.
Corrected integral:
A = 2π∫[0,4] (3√x)√(1 + 9/4x) dx
By evaluating this integral correctly, you should be able to find the correct answer.
To find the area of the surface of revolution, you need to use the formula:
A = 2π ∫[a,b] (y √(1 + (dy/dx)^2)) dx
In this case, the curve is y = 3√(x), so we need to find dy/dx.
dy/dx = (3/2√(x))
Now, we can substitute these values into the formula:
A = 2π ∫[0,4] (3√(x) √(1 + (3/2√(x))^2)) dx
Simplifying further:
A = 2π ∫[0,4] (3x √(1 + (9/4x))) dx
Now, substitute u = 1 + (9/4x). Solving for x, we get x = (9/(4(u-1)).
Differentiating both sides with respect to u gives dx = (-9/4(u-1)^2) du.
Now, we can substitute u and dx in terms of du:
A = 2π ∫[u=1 to u=10/9] (3(9/(4(u-1))) √(u)) (-9/4(u - 1)^2) du
Simplifying further:
A = -2π (27/2) ∫[u=1 to u=10/9] (1/((u - 1)^(3/2))) du
Finally, integrating:
A = -2π (27/2) [(-2(u - 1)^(-1/2))] [u=1 to u=10/9]
A = π (27/4) [(2/√(10/9 - 1)) - (2/√(0))]
Since √(0) is undefined, we need to find the limit as u approaches 1. Taking the limit:
(2/√(10/9 - 1)) = (2/√(1/9))
(2/√(1/9)) = 2/((1/3)) = 6
Therefore:
A = π (27/4) (6 - 0)
A = π (27/4) (6)
A = 162π
So the correct answer is 162π.