Find the surface area generated when
y = (x^3/12) + (1/x), from x=1 to x=2 is rotated around the x-axis.
I visualize an infinite number of "rings"
radius of ring = (x^3)/12 + 1/x
circumference = 2π((x^3)/12 + 1/x)
surface area = 2π∫( (x^3)/12 + 1/x) dx from 1 to 2
= 2π [ (x^4)/48 + lnx] from 1 to 2
= 2π (1/3 + ln 2 - 1/48 - ln1)
= 2π(5/16 + ln 2
= π(5/8 + 2ln 2)
To find the surface area generated by rotating the curve y = (x^3/12) + (1/x) around the x-axis, we can use the formula for surface area of revolution:
S = 2π∫[a,b] y √(1 + (dy/dx)^2) dx
where:
a = 1 (lower limit of x)
b = 2 (upper limit of x)
dy/dx = derivative of y with respect to x
First, let's find the derivative of y:
dy/dx = d/dx[(x^3/12) + (1/x)]
= (3x^2)/12 - (1/x^2)
= x^2/4 - 1/x^2
Now we can substitute these values into the formula:
S = 2π∫[1,2] [(x^3/12) + (1/x)] √(1 + (x^2/4 - 1/x^2)^2) dx
Let's simplify the expression inside the square root:
(1 + (x^2/4 - 1/x^2)^2)
= (1 + (x^4/16) - (2/x^2) + (1/x^4))
= (x^4 + 16x^2 - 32 + 16/x^2 + 1/x^4)/16
= (x^4 + 16x^2 + 16/x^2 + 1/x^4 - 32)/16
Now we can substitute this expression back into the surface area formula:
S = 2π∫[1,2] [(x^3/12) + (1/x)] √((x^4 + 16x^2 + 16/x^2 + 1/x^4 - 32)/16) dx
To calculate the integral, you can use numerical methods or a computer algebra system.
To find the surface area generated when the given curve is rotated around the x-axis, we can use the method of cylindrical shells. The formula for the surface area of a single cylindrical shell is given by:
dA = 2πrh dx,
where r is the radius of the shell, h is its height, and dx is an infinitesimally small width in the x-direction.
To apply this formula, we need to express the curve y = (x^3/12) + (1/x) in terms of x. First, let's find the equation of the curve when rotated around the x-axis.
To rotate the curve around the x-axis, we can create a function representing the radius of each shell. In this case, the radius, r, is equal to the y-value of the curve at each point. So we have:
r = (x^3/12) + (1/x).
Next, we determine the height, h, of each shell. Since we are rotating around the x-axis, the height of each shell is given by the differential dy. So, we have:
dh = dy = dx * ((dx/dy) / (dy/dx)) = dx * (1 / (d/dx((x^3/12) + (1/x)))).
Now, differentiate the equation y = (x^3/12) + (1/x) with respect to x to find dy/dx:
dy/dx = d/dx((x^3/12) + (1/x))
= (3x^2/12) - (1/x^2)
= (x^2/4) - (1/x^2).
Substituting this value into the expression for dh, we have:
dh = dx * (1 / ((x^2/4) - (1/x^2))).
Now, we have the radius (r) and the height (h) of each shell in terms of x. We can use these to calculate the surface area of each shell.
dA = 2πrh dx
= 2π * ((x^3/12) + (1/x)) * (1 / ((x^2/4) - (1/x^2))) * dx.
The total surface area (A) is obtained by integrating the above expression over the given range of x values (from x = 1 to x = 2):
A = ∫(1 to 2) 2π * ((x^3/12) + (1/x)) * (1 / ((x^2/4) - (1/x^2))) dx.
Evaluating this integral will give the surface area generated when y = (x^3/12) + (1/x) is rotated around the x-axis from x = 1 to x = 2.