RaySix

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Volume of Solid of Revolution- Cylindrical Shells

Created by @nathanedwards

at October 31st 2023, 5:16:05 pm.

AP_Calculus_AB-Questions

#calculus

#solid-of-revolution

#cylindrical-shells

Question:

A region in the xy-plane is defined by the curve $y=2x^2$ and the lines $x=0$ and $x=2$ . The region is bounded by the x-axis and the curve. The region is revolved around the x-axis, creating a solid of revolution. Find the volume of the solid.

Answer:

To find the volume of the solid, we will use the method of cylindrical shells. The volume of each cylindrical shell is given by the formula:

V = 2\pi \int_a^b r(x) h(x) dx

Where $r(x)$ is the radius of the shell at a given x-coordinate and $h(x)$ represents the height of the shell. In this case, $r(x)$ is equal to x (the distance from the x-axis to the curve), and $h(x)$ is equal to $2\pi x$ (the circumference of the shell).

We need to find the limits of integration (a and b). The region is bounded by the x-axis and the curve, so the limits of integration are 0 and 2.

Now let's calculate the volume:

V = 2\pi \int_0^2 x \cdot 2\pi x^2 dx = 4\pi^2 \int_0^2 x^3 dx

Using the power rule of integration, we find:

V = 4\pi^2 \left[\frac{1}{4}x^4\right]_0^2

Evaluating at the limits of integration, we get:

V = 4\pi^2 \left(\frac{1}{4}(2)^4 - \frac{1}{4}(0)^4\right)

Simplifying, we have:

V = 4\pi^2 \left(\frac{16}{4}\right) = 4\pi^2 \cdot 4 = 16\pi^2

Therefore, the volume of the solid of revolution is 16π^2 cubic units.