Find the area of the surface generated by revolving the given curve about the y-axis.
8xy^2=2y^6+1 , 1<=y<=2.
thank you so much.. :)
You need x as a function of y to do the integration.
In this case, x(y) = 1/(8y^2) + (y^4)/4
To get the surface area, evaluate:
Integral of 2 pi x(y) dy
(y = 1 to 2)
The indefinite integral is
2 pi [-1/(24y^3) + y^5/20]
= pi[(y^5)/10 - 1/(24y^3)]
Evaluate it at y=2 and y=1, and take the difference. Check my math.
To find the area of the surface generated by revolving the given curve about the y-axis, you can use the method of cylindrical shells.
Step 1: Solve the equation for x in terms of y:
8xy^2 = 2y^6 + 1
Divide both sides by 8y^2:
x = (2y^6 + 1) / (8y^2)
Simplify the equation:
x = (y^4 + 1) / (4y^2)
Step 2: Set up the integral for the surface area using the formula:
A = 2π ∫[a, b] (x(y) * h(y)) dy
In this case, the limits of integration are y = 1 and y = 2, as given.
Step 3: Determine the height function, h(y):
The height function represents the distance between the curve and the axis of rotation, which is the y-axis in this case. Since we're rotating around the y-axis, the height function is simply x(y).
Step 4: Calculate the integral:
A = 2π ∫[1, 2] ((y^4 + 1) / (4y^2)) * dy
Simplify the equation:
A = π/2 ∫[1, 2] (y^2 + 1/y^2) dy
Step 5: Evaluate the integral:
A = π/2 ∫[1, 2] (y^2 + 1/y^2) dy
= π/2 [(1/3)y^3 - (1/y)] |[1, 2]
= π/2 [(1/3)(2^3) - (1/2) - (1/3)(1^3) + (1/1)]
= π/2 [8/3 - 1/2 - 1/3 + 1]
= π/2 [23/6 - 1/2]
= π/2 [23/6 - 3/6]
= π/2 (20/6)
= (10π/6)
= (5π/3) square units
Therefore, the area of the surface generated by revolving the given curve about the y-axis is (5π/3) square units.
To find the area of the surface generated by revolving the given curve about the y-axis, we will use the method of cylindrical shells.
First, let's rewrite the equation of the curve in terms of x:
8xy^2 = 2y^6 + 1
Divide both sides by y^2:
8x = 2y^4 + y^-2
Rearrange the equation to isolate x:
8x = 2y^4 + 1/y^2
Multiply both sides by y^2 to eliminate the fraction:
8xy^2 = 2y^6 + 1
Now, let's find the derivative of both sides of the equation with respect to y:
d/dy (8xy^2) = d/dy (2y^6 + 1)
8x * d(y^2)/dy = d(2y^6)/dy
8x * 2y = 12y^5
16xy = 12y^5
Divide both sides by 4y:
4x = 3y^4
Solve for x:
x = (3y^4) / 4
Now, we can integrate to find the area of the surface.
The formula to calculate the surface area using cylindrical shells is:
A = 2π ∫[a,b] y * f(y) * dx
where f(y) is the radius of the cylindrical shell at a given y-value.
In this case, the limits of integration are 1 <= y <= 2.
The radius, f(y), is the distance from the axis of revolution (y-axis) to the curve. So, we can use the equation we derived earlier for x:
f(y) = (3y^4) / 4
Now, we can substitute the values into the formula and integrate:
A = 2π ∫[1,2] y * (3y^4) / 4 dy
A = π/2 ∫[1,2] 3y^5 dy
Integrate:
A = π/2 [ (3/6) y^6 ] [1,2]
A = π/2 [ (3/6) 2^6 - (3/6) 1^6 ]
A = π/2 [ 192/6 - 3/6 ]
A = π/2 [ 32 - 1/2 ]
A = π/2 * 63/2
A = 99π/4
So, the area of the surface generated by revolving the given curve about the y-axis is 99π/4.