Find the volume of a solid obtained by revolving the region bounded by y^2-x^2=1 and the y=2 about the x-axis.
y^2-x^2=1 or
x^2 - y^2 = -1 <----- vertical hyperbola, vertices at (0, ± 1)
y=2 intersect the upper part of the hyperbola at (-√3,2) and (√3,2)
because of the symmetry we could just go from 0 to √3, then double our answer.
Volume = 2π∫ (2^2 - y^2) dx from x = 0 to √3
= 2π∫(4 - 1 - x^2) dx from 0 to √3
= 2π[3x - x^3/3] from 0 to √3
= 2π(3√3 - 3√3/3 - 0)
= 4π√3
check my arithmetic , should have written it out on paper.
To find the volume of the solid obtained by revolving the region bounded by the curves, we can use the method of cylindrical shells.
Step 1: Sketch the region bounded by the curves.
Start by graphing the equation y^2 - x^2 = 1.
The graph represents a hyperbola centered at the origin, with branches opening upwards and downwards. The equation y = 2 represents a horizontal line.
Step 2: Identify the limits of integration.
To find the limits of integration, we need to determine the x-values where the curves intersect. Let's rearrange the equation y^2 - x^2 = 1 to solve for x first.
y^2 - x^2 = 1
x^2 = y^2 - 1
x = ±√(y^2 - 1)
Since the region is symmetric about the y-axis, we can focus on the positive x-values.
The curves intersect when y = 2 and y = √2. Therefore, the limits of integration are y = √2 to y = 2.
Step 3: Set up the integral using the formula for the volume of a solid of revolution.
For the cylindrical shells method, the formula for the volume of a solid of revolution is:
V = 2π ∫ [radius * height] dy
In this case, the radius is the distance from the y-axis to the curve, which is x = √(y^2 - 1).
The height is the difference in y-values, which is 2 - √2.
So, the integral to find the volume becomes:
V = 2π ∫ [√(y^2 - 1) * (2 - √2)] dy
Integrated from √2 to 2.
Step 4: Evaluate the integral to find the volume.
You can now evaluate the integral using standard integral techniques or a computer algebra system.
After evaluating the integral, you will find the volume of the solid of revolution.