Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.
y = (x³/6) + (1/2x), 1≤ x ≤ 2
I assume (1/2x) is meant to be 1/(2x), so
y(x) = (x³/6) + 1/(2x), x1≤ x ≤ x2 ,
where x1=1, x2=2.
You are probably aware of the formula for surface of revolution about the x-axis generated by a function y(x) from x=x1 to x=x2:
S=2π∫y(x)√(1+y'(x)^2)dx
We will now calculate each component individually and then do the integration.
y(x)=x³/6+1/(2x) ....(given)
By differentiation with respect to x, we get:
y'(x)=x²-1/(2x²)
Now we proceed to evaluate the expression inside the square-root radical:
√(1+y'(x)²)
=√(1+(x²-1/(2x²))^2)
=√(1+((x^4-1)/(2x²)))
find common denominator and add:
=√((x^8+2x^4+1)/(4x^4))
=(x^4+1)/(2x²)
Now we are ready to put everything together:
Area of revolution
=2π∫(x³/6+1/(2x))*((x^4+1)/(2x²) dx from x1 to x2
=2π∫((x^4+3)/(6x))*((x^4+1)/(2x²) dx
=2π∫((x^4+3)(x^4+1)/(12x³)) dx
=2π∫((x^8+4x^4+3)/(12x³)) dx
=(π/6)∫((x^5+4x+3/x³)) dx
=(π/6)(x^6/6+2x²-3/(2x²))
Evaluate integral between x1=1 and x2=2 gives 47π/16 (=9.2 approximately).
As a rough check, we calculate the function at x=(x1+x2)/2=1.5 and estimate the area:
Aapprox=2πf(1.5)*radic;(1+f'(1.5)^2)
=2π(0.9)√(1+0.9²)*(x2-x1)
=7.6, which is not too far from our calculations above.
In this case, a better check is by using simpson's rule for numerical integration, given by:
∫g(x) from a to b
=(b-a)/6*(g(a)+4g((a+b)/2)+g(b))
and in the present case,
g(x)=2πy(x)sqrt(1+y'(x))
so
A=((2-1)/6)*(g(1)+4g(1.5)+g(2))
=(1/6)(4π/3+4*2.41π+6.73π)
=9.28, much closer to 9.23 by integration.
To set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis, we can use the formula for the surface area of revolution:
S = 2π ∫[a,b] y √(1 + (dy/dx)²) dx
First, let's find the derivative of y with respect to x:
dy/dx = (d/dx)(x³/6 + 1/(2x))
= (3x²/6) - (1/(2x²))
= (x²/2) - (1/(2x²))
The integrand in the surface area formula is y √(1 + (dy/dx)²), so we can substitute the expressions for y and dy/dx into this formula:
y = (x³/6) + (1/2x)
dy/dx = (x²/2) - (1/(2x²))
Now, we can set up the definite integral with the given limits of integration:
S = 2π ∫[1,2] ((x³/6) + (1/2x)) √(1 + ((x²/2) - (1/(2x²)))²) dx
To evaluate this integral, we can simplify the expression inside the square root and then evaluate the integral numerically or symbolically.
To set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis, we need to follow these steps:
Step 1: Determine the formula for the infinitesimal surface area element.
The infinitesimal surface area element can be approximated by the arc length of a differential segment of the curve multiplied by the circumference of the circle formed by revolving that segment around the x-axis.
Step 2: Find the formula for the arc length of the curve.
For the given curve, we can find the arc length by integrating the square root of the sum of the squares of the derivatives of x and y with respect to x, which is represented by the formula:
ds = sqrt(1 + (dy/dx)^2) dx
Step 3: Find the circumference of the circle formed by revolving the curve.
Since we are revolving the curve about the x-axis, the circumference of the circle formed by each infinitesimal segment is given by 2πy.
Step 4: Set up the integral for the surface area.
To get the total surface area, we need to integrate the infinitesimal surface area element over the given range [1, 2].
Step 5: Evaluate the definite integral to find the area.
Once you have set up the integral, you can evaluate it using integration techniques and the given function.
Now let's apply these steps to the given problem:
Step 1: The formula for the infinitesimal surface area element is:
dA = 2πy * ds
Step 2: Finding the arc length:
ds = sqrt(1 + (dy/dx)^2) dx
dy/dx = d(x³/6 + 1/(2x))/dx = (x²/2)-(1/(2x²))
ds = sqrt(1 + ((x²/2)-(1/(2x²)))^2) dx
Step 3: Finding the circumference of the circle:
C = 2πy = 2π((x³/6) + (1/2x))
Step 4: Setting up the integral:
The definite integral for the surface area is:
A = ∫(2πy * ds) from 1 to 2
Substituting the values for y and ds:
A = ∫(2π((x³/6) + (1/2x)) * sqrt(1 + ((x²/2)-(1/(2x²)))^2)) dx from 1 to 2
Step 5: Evaluating the definite integral:
To evaluate the integral, you can either solve it manually using integration techniques or use software or calculators capable of performing definite integrals.