Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 2/x, y = 0, x = 1, x = 3; about y = −1
V=
To find the volume, we use the method of cylindrical shells, which gives us the formula:
V = ∫[a,b] 2πrh dx
where r is the radius and h is the height.
In this case:
- The region is bounded by y = 2/x, y = 0, x = 1, and x = 3.
- We have to rotate it around y = -1.
- So, the radius r = y+1 = 2/x + 1 and the height h = x.
Now, we can plug the values into the formula and integrate:
V = ∫[1,3] 2π(2/x + 1)x dx
= 2π ∫[1,3] (x(2/x) + x) dx
= 2π ∫[1,3] (2 + x) dx
Now, we integrate:
V = 2π [2x + (1/2)x^2] evaluated from 1 to 3
= 2π [(2(3) + (1/2)(3^2)) - (2(1) + (1/2)(1^2))]
= 2π [(6 + 4.5) - (2 + 0.5)]
= 2π (10.5 - 2.5)
= 2π (8)
V = 16π cubic units.
To find the volume of the solid obtained by rotating the region bounded by the curves y = 2/x, y = 0, x = 1, and x = 3 about the line y = −1, we will use the method of cylindrical shells.
Let's begin by visualizing the region bounded by the curves and the line of rotation:
First, let's find the intersection points of the curves:
Setting y = 0 in y = 2/x, we have:
0 = 2/x
This implies x = 2.
So the region is bounded by the x-axis, x = 1, x = 3, and the curve y = 2/x.
Now, let's consider a typical shell within this region with radius r and height h. We will consider the horizontal shells since we are rotating about the vertical line y = −1.
The radius of the shell at any given height y is the distance from the line of rotation y = −1 to the curve y = 2/x. Since y = 2/x, we can rewrite it as x = 2/y. Thus, the radius r is given by r = 2/y − (−1) = 2/y + 1.
The height of the shell h is the difference between the upper and lower y-values of the region. In this case, the upper y-value is y = 0 (the x-axis) and the lower y-value is given by the curve y = 2/x. So, h = 0 − (2/x) = −2/x.
The volume of a single shell is given by the formula: dV = 2πrh dh.
To calculate the total volume, we integrate dV with respect to y from y = 0 to y = ∞ (the region extends to infinity vertically because the curve y = 2/x approaches but never touches the y-axis).
So, the volume V is:
V = ∫[0,∞] (2πrh) dy
= ∫[0,∞] (2π((2/y + 1)(−2/x))) dy
= −4π ∫[0,∞] ((1/y)x) dy
Now, let's integrate with respect to y:
V = −4π ∫[0,∞] (x/y) dy
To simplify the integration, we can change the order of integration by considering x as a function of y. From y = 2/x, we know that x = 2/y.
V = −4π ∫[0,∞] ((2/y)/y) dy
= −8π ∫[0,∞] (1/y^2) dy
Now, we can solve the integration:
V = −8π (−1/y)|[0,∞]
= 8π (1/∞ − 1/0)
= 8π (0 − 1/0)
= 8π (−∞)
= −∞
So, the volume of the solid obtained by rotating the region bounded by the curves y = 2/x, y = 0, x = 1, and x = 3 about the line y = −1 is undefined.
To find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, we can use the method of cylindrical shells.
First, we need to set up the integral to calculate the volume.
Let's consider a small vertical strip in the region bounded by the curves. The width of this strip is dx, the height is y, and the length (circumference) of the strip when rotated about the line y = -1 is 2π(y + 1). Hence, the volume of this strip is approximately equal to (2π(y + 1)) * y * dx.
To find the limits of integration for x, we can solve the equation of the two curves for the x-coordinate of their intersection points:
2/x = 0 (setting y = 0)
x = 2/x
x^2 = 2
x = ±√2
Since the region is bounded by x = 1 and x = 3, our limits of integration for x will be √2 and 3.
Now, let's set up the integral to calculate the volume:
V = ∫(√2 to 3) [(2π(y + 1)) * y] dx
To express y in terms of x, we rearrange the equation y = 2/x to get: x = 2/y.
Substituting x = 2/y into the integral, we have:
V = ∫(√2 to 3) [(2π(2/y + 1)) * y] dx
Simplifying the expression inside the integral, we get:
V = ∫(√2 to 3) [(4π/y + 2π) * y] dx
= ∫(√2 to 3) (4π + 2πy) dx
= ∫(√2 to 3) (4πx + 2πx^2) dx
Integrating with respect to x, we have:
V = [ 2πx^2 + (2/3)πx^3 ] evaluated from √2 to 3
= [ 2π(3)^2 + (2/3)π(3)^3 ] - [ 2π(√2)^2 + (2/3)π(√2)^3 ]
= 18π + (18/3)π - 2π - (2/3)π√2
= 12π - (2/3)π√2
Therefore, the volume V of the solid obtained by rotating the region bounded by the given curves about the line y = -1 is 12π - (2/3)π√2.