Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=0, y= \cos(x), x = \frac{\pi}{2}, x = 0 about the axis y= -3
You might need to use the identity: cos^2(x) = \frac{1}{2}(cos(2x) + 1).
To find the volume of the solid obtained by rotating the region bounded by the curves about the axis y = -3, we can use the method of cylindrical shells.
Step 1: Determine the interval of integration
The given curves y = 0, y = cos(x), x = π/2, and x = 0 bound the region of interest. To find the interval of integration, we need to determine the values of x that define the region. From the given curves, we know that x ranges from 0 to π/2.
Step 2: Set up the integral
We can express the volume of a cylindrical shell as V = 2πrhΔx, where r is the distance from the axis of rotation (y = -3), h is the height of the shell (given by the difference between the curves y = 0 and y = cos(x)), and Δx represents an infinitesimally small width along the x-axis.
Since we are rotating about the line y = -3, the distance from the axis of rotation to a point on the curve y = cos(x) is (cos(x) + 3), so r = cos(x) + 3. The height of the shell, h, is given by the difference between the curves y = cos(x) and y = 0, which is simply cos(x). Therefore, the volume element is dV = 2π(cos(x) + 3)cos(x)Δx.
Step 3: Evaluate the integral
To find the total volume, we integrate the volume element over the interval [0, π/2].
V = ∫[0,π/2] 2π(cos(x) + 3)cos(x) dx
Now, we can simplify the integral by expanding and applying the given identity: cos^2(x) = 1/2(cos(2x) + 1).
V = 2π ∫[0,π/2] [(cos(x)cos(x) + 3cos(x)] dx
= 2π ∫[0,π/2] [1/2(cos(2x) + 1) + 3cos(x)] dx
= π [∫[0,π/2] cos(2x) dx + ∫[0,π/2] dx + 3∫[0,π/2] cos(x) dx]
The integral of cos(2x) can be evaluated using the substitution u = 2x, du = 2dx:
V = π [1/2 ∫[0,π] cos(u) du + ∫[0,π/2] dx + 3∫[0,π/2] cos(x) dx]
= π [1/2 sin(u)|[0,π] + x|[0,π/2] + 3 sin(x)|[0,π/2]]
= π [1/2 (sin(π) - sin(0)) + (π/2 - 0) + 3(sin(π/2) - sin(0))]
Simplifying further:
V = π [1/2 (0 - 0) + (π/2) + 3(1 - 0)]
= π [π/2 + 3]
= π^2/2 + 3π
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = 0, y = cos(x), x = π/2, and x = 0 about the axis y = -3 is π^2/2 + 3π cubic units.