The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method.

Question

y² − x² = 1,    y = 2;    about the x-axis

Answer 1

Intersection points x-coordinate: 2² - x² = 1; x² = 3, x = ±√3.

Using disk method we have V = π∫_-√3^√3₍₂² - (1 +x²))dx = π∫_-√3^√3(3 - x²)dx = 2π∫₀^√3(3 - x²)dx = 2π(3x - x³/3)₀^√3= 2π(3√3 - √3) = 4π√3 ≈ 21.77

Answer 2

Graphing the equation you've given, it appears to be a hyperbola with y = 1 when x = 0, and the upper bound is y = 2. (We can ignore the negative side since it's outside of the bounded region.)

Essentially, the volume of the solid will be the sum of all the areas of the discs whose center points are on the x-axis and whose radii are the positive x-values for every y.

To calculate an area of an individual disc at a given y-value, we need to know the x-value for that y, and substitute that into the formula for the volume of a circle. We must solve the equation in terms of x: we get x = √(y² - 1), meaning that the area of a disc at a given y will be π(√(y² - 1))², or π(y² - 1).

To calculate the successive sums of these discs, we can take the integral of π(y² - 1) dy, with limits of integration from 1 (the lower bound since that's where the volume ends) to the given upper bound of 2.

Solving this definite integral on a calculator, we get that the volume of the solid is 4π/3, or 4.189 rounded to three decimal places.