Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 4 − 1/2x, y = 0, x = 0, x = 1; about the x-axis
V = ????
think of discs with radius y, thickness dx, so
v = ∫[0,1] πy^2 dx
= π∫[0,1] (4-x/2)^2 dx
a nice easy integral. You should get 169/12 π
To find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis, we can use the method of cylindrical shells.
The formula for finding the volume using cylindrical shells is:
V = 2π ∫(a to b) (radius) (height) dx
Where a and b are the x-coordinates of the region being rotated, the radius is the perpendicular distance from the axis of rotation to the curve, and the height is the differential length along the x-axis.
In this case, the curve y = 4 - (1/2)x intersects the x-axis at y = 0. The x-values that bound the region are x = 0 and x = 1.
The radius can be found as the distance from the x-axis to the curve, which is y.
So, the radius is y = 4 - (1/2)x and the height is dx.
The volume V can be written as:
V = 2π ∫(0 to 1) (4 - (1/2)x) dx
Now, we can integrate to find the volume:
V = 2π ∫(0 to 1) (4x - (1/2)x^2) dx
V = 2π [(2x^2 - (1/6)x^3) from 0 to 1]
V = 2π [(2 - (1/6)) - (0 - 0)]
V = 2π [(2/3)]
V = (4/3)π
Therefore, the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis is (4/3)π cubic units.
To find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis, we can use the method of cylindrical shells.
First, let's find the height of the solid. Since the rotation is happening about the x-axis, the height of each cylindrical shell will be determined by the distance between the two curves at each x-value.
The two curves are y = 4 - (1/2)x and y = 0. To find the height, we subtract the lower curve from the upper curve: (4 - (1/2)x) - 0 = 4 - (1/2)x.
Next, let's find the radius of each cylindrical shell. The radius will be the x-value since the x-axis is the axis of rotation.
The region is bounded by x = 0 and x = 1. Therefore, the radius will vary from 0 to 1.
Now, we can set up the integral to calculate the volume using the formula for the volume of a cylindrical shell:
V = ∫[lower bound, upper bound] 2πrh dx
In this case, the lower bound is 0 and the upper bound is 1.
V = ∫[0, 1] 2πx(4 - (1/2)x) dx
Now, we can integrate this expression to find the volume V:
V = 2π ∫[0, 1] (4x - (1/2)x^2) dx
To find the antiderivative of each term, we use the power rule of integration:
V = 2π [(2x^2/2) - ((1/2)(1/3)x^3)] evaluated from 0 to 1
Simplifying:
V = 2π [x^2 - (1/6)x^3] evaluated from 0 to 1
Now, substitute the upper bound (1) and the lower bound (0) into the expression:
V = 2π [(1^2 - (1/6)(1)^3) - (0^2 - (1/6)(0)^3)]
V = 2π [(1 - (1/6) - (0 - 0)]
V = 2π (1 - 1/6)
V = 2π (5/6)
Therefore, the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis is (10π/6) or (5π/3) cubic units.