How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y = 2 x^4#, y = 0, x = 1 revolved about the x=2?

Question

Charles Anctil · Accepted Answer

See the answer below:

Ezra Adams · Answer

undefined To use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by $ y = 2x^4 $, $ y = 0 $, $ x = 1 $ revolved about $ x = 2 $, follow these steps:

1. Determine the limits of integration. In this case, the region is bounded by $ y = 2x^4 $, $ y = 0 $, and $ x = 1 $. The rotation is about the vertical line $ x = 2 $, so the limits of integration will be from $ x = 1 $ to $ x = 2 $.

2. Set up the integral for the volume using the formula for cylindrical shells:

$$ V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx $$

where $ h(x) $ is the height of the shell and $ a $ and $ b $ are the lower and upper limits of integration, respectively.

3. Express the height of the shell, $ h(x) $, as the difference between the outer radius and the inner radius. The outer radius is the distance from the axis of rotation ( $ x = 2 $ in this case) to the curve $ y = 2x^4 $, which is $ 2 - x $. The inner radius is the distance from the axis of rotation to the line $ x = 1 $, which is $ 2 - 1 = 1 $.

4. Calculate the integral:

$$ V = 2\pi \int_{1}^{2} x \cdot (2 - x - 1) \, dx $$

5. Integrate the expression:

$$ V = 2\pi \int_{1}^{2} (2x - x^2 - x) \, dx $$

$$ V = 2\pi \left[ \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^2}{2} \right]_{1}^{2} $$

$$ V = 2\pi \left[ \left(2 - \frac{8}{3} - 2\right) - \left(\frac{1}{2} - \frac{1}{3} - \frac{1}{2}\right) \right] $$

$$ V = 2\pi \left[ \frac{4}{3} - \left(\frac{1}{6}\right) \right] $$

$$ V = 2\pi \left( \frac{7}{6} \right) $$

$$ V = \frac{7\pi}{3} $$

Therefore, the volume of the solid obtained by rotating the region about $ x = 2 $ using the method of cylindrical shells is $ \frac{7\pi}{3} $.