Find the volume of the solid obtained by rotating the region enclosed by the curves y=1x and y=3−x about the x-axis.
I assume you meant y = 1/x
If so, then the curves intersect at
((3+√5)/2,(3-√5)/2) and ((3-√5)/2,(3+√5)/2)
call these points (a,b) and (b,a)
So, using discs (washers) of thickness dx, we have
v = ∫[a,b] π(R^2-r^2) dx
where R=3-x and r=1/x
= ∫[a,b] π((3-x)^2-1/x^2) dx = 5π√5/3
Or, using shells of thickness dy,
v = ∫[a,b] 2πrh dy
where r=y and h=(3-y)-(1/y)
= ∫[a,b] 2πy((3-y)-(1/y)) dy = 5π√5/3
ahem. For shells, it is ∫[b,a]
To find the volume of the solid obtained by rotating the region enclosed by the curves y=1x and y=3−x about the x-axis, we can use the method of cylindrical shells.
First, let's find the points of intersection between the curves y=1x and y=3−x. To do this, we can set the two equations equal to each other:
1x = 3−x
Simplifying, we have:
2x = 3
Dividing both sides by 2, we get:
x = 3/2
So the points of intersection are (3/2, 3/2).
Now, we need to find the limits of integration. Since we are rotating about the x-axis, the lower limit of integration is the x-coordinate of the leftmost point of intersection, which is 0, and the upper limit is the x-coordinate of the rightmost point of intersection, which is 3/2.
Next, we need to express the curves y=1x and y=3−x in terms of x. Both curves are already in terms of x, so we don't need to make any changes.
The formula for the volume of the solid obtained by rotating the region bounded by the curves y=f(x) and y=g(x) about the x-axis is given by:
V = ∫[a,b] 2πx(f(x)−g(x)) dx,
where [a, b] is the interval of integration.
Applying this formula to our problem, we have:
V = ∫[0,3/2] 2πx((3−x)−(1x)) dx
Simplifying this expression:
V = ∫[0,3/2] 2πx(3−2x) dx
Expanding the expression further:
V = ∫[0,3/2] (6πx−4πx^2) dx
Integrating term by term:
V = (3πx^2−4πx^3/3)|[0,3/2]
Evaluating the integral at the upper and lower limits:
V = (3π(3/2)^2−4π(3/2)^3/3) − (3π(0)^2−4π(0)^3/3)
Calculating the values:
V = (9π/4−18π/8) − (0−0)
Simplifying:
V = (9π−9π/4) − 0
Which gives us:
V = 27π/4 cubic units
Therefore, the volume of the solid obtained by rotating the region enclosed by the curves y=1x and y=3−x about the x-axis is 27π/4 cubic units.