Find the volume formed by rotating the region enclosed by:
x=2y and y^3=x with y greater than or equal to 0 about the y-axis
There are two regions between the curves, from (0,0) to (2 sqrt 2, sqrt 2) and the same in the third quadrant,
So do the integral in quadrant 1 and double.
We could do little vertical cylinders and integrate over x or horizontal rings and integrate over y.
For the rings;
integral of
dy pi (x outer^2 - x inner^2) from y =0 to y = sqrt 2
where x outer = 2y and x inner = y^3
pi dy (4 y^2 -y^6)
then
pi [ (4/3)y^3 -(1/7)y^7 ]
pi [ (4/3) 2^(3/2) - (1/7) 2^(7/2) ]
check my numbers !!!
I often found that students were more comfortable and familiar with rotations around the x-axis.
Notice we could just switch the x and y variables, and then rotate around the x-axis.
All shapes and volumes would be retained.
Then volume
= 2pi[integral] (4x^2 - x^6)dx from 0 to √2
the actual work and calculations would be same as Damon showed you.
Just remember to double Damon's final answer because we have two identical solids.
To find the volume formed by rotating the region enclosed by the curves x = 2y and y^3 = x about the y-axis, we can use the method of cylindrical shells.
First, let's sketch the region to get a visual understanding of the problem.
The curves x = 2y and y^3 = x intersect at (0, 0) and (8, 2). The region enclosed is a bounded region between these two points.
Next, let's express both curves in terms of y, so we can determine the limits of integration.
From x = 2y, we have y = x/2.
From y^3 = x, we have x = y^3.
Thus, the limits of integration for y are from 0 to 2.
Now, we can set up the integral for the volume of the region using cylindrical shells:
V = ∫(2πy)(x) dy
To express x in terms of y, substitute x = y^3 into the integral:
V = ∫(2πy)(y^3) dy
Simplifying, we have:
V = 2π ∫(y^4) dy
Integrating, we get:
V = 2π (y^5/5)
To evaluate the definite integral, substitute the limits of integration:
V = 2π [(2^5/5) - (0^5/5)]
V = 2π (32/5)
V = 64π/5
So, the volume formed by rotating the region enclosed by x = 2y and y^3 = x about the y-axis is 64π/5 cubic units.
To find the volume formed by rotating a region about the y-axis, we can use the method of cylindrical shells. The formula for the volume of a cylindrical shell is given by:
V = 2π ∫[a,b] (x * h) dx
where a and b are the limits of integration, x is the distance from the axis (in this case, the x-coordinate of the curve), and h is the height of the shell (in this case, the difference in y-coordinates of the curve).
Now let's find the limits of integration a and b.
From the equations x = 2y and y^3 = x, we can rewrite the first equation as y = x/2, and substitute this into the second equation:
(y^3) = x
(y^3) = (x/2)^3
y = (x/2)^(1/3)
Now let's find the limits of integration a and b by setting the two equations equal to each other:
(x/2)^(1/3) = x/2
x^(1/3) = x/2
2x^(1/3) = x
2 = x^(2/3)
x = 8
So the limits of integration are a = 0 and b = 8.
Now let's find the height h of the cylindrical shell. It is the difference between the y-coordinates of the curve:
h = (x/2)^(1/3) - 0 = (x/2)^(1/3)
Substituting these values into the volume formula, we have:
V = 2π ∫[0,8] (x * (x/2)^(1/3)) dx
Integrating this expression will give us the volume.