Find the volume of the solid obtained by rotating the region bounded by the curves:
y = x^2 and x = y^2 about the line y = 1.
I received the answer pi/30, using the outer radius of (1 - x^2) and the inner radius of (1 - x^(1/2)). Is this correct?
Scratch that, I received the answer of 11pi/30, I made a calculation error that first time. Is this correct?
As usual, you can use shells or discs (washers). I think shells is easier here.
Each shell has the line y=1 as its axis, and height the distance between the curves. We integrate on y, since the shells have thickness dy
v = ∫2πrh dy
where r = (1-y) and h = √y-y^2
v = 2π∫[0,1] (1-y)(√y-y^2) dy
= 11π/30
If you want to use washers, then integrate along x, and we have
v = ∫π(R^2-r^2) dx
where R = 1-x^2 and r=1-√x
v = π∫[0,1] (1-x^2)^2 - (1-√x)^2 dx
= 11π/30
Check these solutions against your work, to see where you went wrong.
Thank you!
To find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the line y = 1, you can use the method of cylindrical shells.
First, let's plot the curves and the axis of rotation to visualize the solid:
Graph:
---------------
| x-axis
|
| y-axis
|
| ^
| |
|-------------|----- y = 1
| |
| |
| |
| |
The solid is formed by rotating the region between the curves y = x^2 and x = y^2 about the line y = 1.
To set up the integral, consider an infinitesimally small vertical strip of width dx at x. This strip, when rotated, forms a cylindrical shell. We want to find the volume of all these shells and integrate it over the entire region.
The radius of each cylindrical shell is the distance from the curve x = y^2 to the line y = 1, which is (1 - x^(1/2)). The height of each shell is the difference in y-values between the curves y = x^2 and y = 1, which is (x^2 - 1).
The volume of each cylindrical shell is given by the formula:
V = 2π * radius * height * dx
So, the total volume V can be obtained by integrating this expression over the region (x = 0 to x = 1):
V = ∫[0 to 1] 2π * (1 - x^(1/2)) * (x^2 - 1) dx
Now let's evaluate this integral to find the volume:
V = 2π * ∫[0 to 1] (x^2 - x^(1/2) - x^(5/2) + x^(3/2)) dx
Integrating term by term:
V = 2π * (x^3/3 - (2/3)x^(3/2) - (2/7)x^(7/2) + (2/5)x^(5/2)) evaluated from x = 0 to x = 1
V = 2π * [(1/3) - (2/3) - (2/7) + (2/5)]
V = 2π * (1 - 2/3 - 2/7 + 2/5)
V = 2π * (15/105 - 70/105 - 30/105 + 42/105)
V = 2π * (15 - 70 - 30 + 42)/105
V = 2π * (-43)/105
V = -86π/105
So, the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and x = y^2 about the line y = 1 is -86π/105 or approximately -0.82π.