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Volume of Solid of Revolution for y = x^2

Created by @nathanedwards
 at October 31st 2023, 7:47:39 pm.
AP_Calculus_AB-Questions
#integration
#mathematics
#volume

Question:

Find the volume of the solid obtained by rotating the region under the curve y = x^2, bounded by the x-axis and the vertical line x = 2, about the x-axis. Provide your answer as a decimal rounded to two decimal places.

Solution:

To find the volume of the solid obtained by revolving the region under the curve y = x^2 about the x-axis, we can use the method of cylindrical shells.

First, let's sketch the given region:

Volume of solids of revolution

The region under the curve y = x^2, bounded by the x-axis and the vertical line x = 2 looks like a right triangle in this case.

Next, we need to determine the height of the cylindrical shell. Since we are rotating the region about the x-axis, the height of the shell will be the y-coordinate of the curve at each x-value.

The equation of the curve is y = x^2, so the height of the shell can be expressed as (x^2).

Now, we need to determine the radius of the cylindrical shell. Since we are rotating about the x-axis, the radius of the shell is the distance from the x-axis to the vertical line x = 2. Therefore, the radius is (2 - x).

The thickness of the cylindrical shell is given by "dx", which represents an infinitesimally small change in x.

The volume of each cylindrical shell can be calculated using the formula:

Volume of shell=2π⋅height⋅radius⋅thickness \text{{Volume of shell}} = 2\pi \cdot \text{{height}} \cdot \text{{radius}} \cdot \text{{thickness}}

Substituting the given values, we have:

Volume of shell=2π⋅(x2)⋅(2−x)⋅dx \text{{Volume of shell}} = 2\pi \cdot (x^2) \cdot (2 - x) \cdot dx

To find the total volume, we need to integrate the volume of each shell from 0 to 2:

Total volume=∫022π⋅(x2)⋅(2−x)⋅dx \text{{Total volume}} = \int_0^2 2\pi \cdot (x^2) \cdot (2 - x) \cdot dx

Simplifying the expression inside the integral, we have:

Total volume=2π∫02(2x2−x3)⋅dx \text{{Total volume}} = 2\pi \int_0^2 (2x^2 - x^3) \cdot dx

Now, let's integrate:

Total volume=2π[2x33−x44]02 \text{{Total volume}} = 2\pi \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_0^2

Evaluating the definite integral, we get:

Total volume=2π(2(23)3−244−(2(03)3−044)) \text{{Total volume}} = 2\pi \left( \frac{2(2^3)}{3} - \frac{2^4}{4} - \left( \frac{2(0^3)}{3} - \frac{0^4}{4} \right) \right)

Simplifying the expression, we find:

Total volume=2π(163−164) \text{{Total volume}} = 2\pi \left( \frac{16}{3} - \frac{16}{4} \right)
Total volume=2π(163−4) \text{{Total volume}} = 2\pi \left( \frac{16}{3} - 4 \right)
Total volume=83π \text{{Total volume}} = \frac{8}{3}\pi

Rounding to two decimal places, the volume is approximately 8.38 cubic units.

Therefore, the volume of the solid obtained by rotating the region under the curve y = x^2, bounded by the x-axis and the vertical line x = 2, about the x-axis is approximately 8.38 cubic units.

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