Find the volume of the solid generated by revolving the following region about the given axis. The region in the first quadrant bounded by the curve y=x^2, below by the x-axis, and on the right by the line x=1, about the line x=-2
Using shells,
v = Int(2πrh dx)[0,1]
where
r = x+2
h = y = x^2
2π*Int((x+2)x^2 dx)[0,1]
2π*Int(x^3 + 2x^2 dx)[0,1]
= 2π(1/4 x^4 + 2/3 x^3)[0,1]
= 2π(1/4 + 2/3) = 11π/6
To find the volume of the solid generated by revolving the given region about the line x = -2, we can use the method of cylindrical shells.
First, let's visualize the region we need to revolve. The region is in the first quadrant bounded by the curve y = x^2, below by the x-axis, and on the right by the line x = 1.
To get started, let's sketch the region:
y |\
| \
________|__\_____________
| | \
| | \
| | \
-2 1 x
Now, in order to use the cylindrical shell method, we need to express the volume of each infinitesimally thin shell. Each shell has a height of dy and radius of (x + 2) (since we are revolving about the line x = -2).
The volume of each cylindrical shell is given by 2πrh, where r is the radius and h is the height. In this case, r = x + 2 and h = dy.
Therefore, the volume of each shell is 2π(x + 2)dy.
To find the total volume, we need to integrate these shells over the range of y-values that correspond to the region we're revolving. In this case, the y-values range from 0 (the x-axis) to 1 (the curve y = x^2).
So, the integral to find the volume is:
V = ∫[0,1] 2π(x + 2)dy
To express x in terms of y, we rearrange the equation y = x^2:
x = √y
Now, we substitute x = √y into the integral:
V = ∫[0,1] 2π(√y + 2)dy
Now we can integrate:
V = 2π ∫[0,1] (√y + 2)dy
V = 2π [ (√y^3)/3 + 2y ] │[0,1]
V = 2π ( [(1^(3/2))/3 + 2(1)] - [(0^(3/2))/3 + 2(0)] )
V = 2π ( (1/3) + 2 )
V = 2π (7/3)
V = (14π/3)
So, the volume of the solid generated by revolving the given region about the line x = -2 is (14π/3) cubic units.