The portion of the ellipse x^2/9+y^2/4=1 with x greater than or equals to 0 is rotated about the y-axis to form a solid S. A hole of radius 1 is drilled through the center of S, along the y-axis. Find the exact volume of the part of S that remains. Show all steps. (Hint: use shells)
To find the volume of the part of solid S that remains after the hole is drilled through it, we can use the method of cylindrical shells.
First, let's visualize the problem. The given equation x^2/9 + y^2/4 = 1 represents an ellipse centered at the origin (0,0) with semi-major axis (a = 3) and semi-minor axis (b = 2). The part of the ellipse with x ≥ 0 is the right half of the ellipse.
When this right half of the ellipse is rotated about the y-axis to form solid S, it creates a solid object that resembles a bottle with a hole drilled through its center along the y-axis. We want to find the volume of the remaining part of S after the hole is drilled.
Now, let's use the method of cylindrical shells to find the volume. Consider a vertical strip of thickness dx at a distance x from the y-axis. When this strip is rotated around the y-axis, it forms a cylindrical shell with height y and a very small thickness dx.
The circumference of this cylindrical shell is given by 2πy, and its height is dx. Therefore, the volume of this cylindrical shell is approximately given by dV = 2πy · dx.
To find y in terms of x, we can rearrange the equation of the ellipse:
x^2/9 + y^2/4 = 1
y^2/4 = 1 - x^2/9
y^2 = 4 - 4x^2/9
y = sqrt(4 - 4x^2/9)
Therefore, the volume of the cylindrical shell can be written as dV = 2π · sqrt(4 - 4x^2/9) · dx.
To find the total volume of the remaining part of S, we integrate dV from x = 0 to x = 3 (the limits of the right half of the ellipse) since the original equation specified x ≥ 0:
V = ∫[0,3] 2π · sqrt(4 - 4x^2/9) · dx
We can simplify the equation inside the square root:
V = ∫[0,3] 2π · sqrt((36 - 4x^2)/9) · dx
V = (2π/3) ∫[0,3] sqrt(36 - 4x^2) · dx
Now, we can use a trigonometric substitution to evaluate this integral. Let x = (3/2)sinθ, then dx = (3/2)cosθdθ:
V = (2π/3) ∫[0,π/2] sqrt(36 - 4(3/2)sinθ)^2 · (3/2)cosθdθ
V = (2π/3) ∫[0,π/2] sqrt(36 - 9sin^2θ) · (3/2)cosθdθ
V = π ∫[0,π/2] sqrt(36cos^2θ + 27sin^2θ) · cosθdθ
Using the identity cos^2θ + sin^2θ = 1, we can simplify further:
V = π ∫[0,π/2] sqrt(36 + 27sin^2θ) · cosθdθ
This integral can be evaluated using trigonometric identities or by using a calculator or a computer algebra system. The result will be the exact volume of the remaining part of solid S after the hole is drilled through it, as requested.