Answer is: option2
\( \pi \int_0^1 \left( [1 + \sin(\pi x)]^2 - x^6 \right) \, dx \)Solution:
To compute the volume of a solid of revolution about the x-axis, we use the washer method:
\[ V = \pi \int_a^b \left[ (f(x))^2 - (g(x))^2 \right] dx \]
Where:
- \( f(x) \) is the outer radius (upper function),
- \( g(x) \) is the inner radius (lower function).
Here:
- Upper function (outer radius): \( f(x) = 1 + \sin(\pi x) \)
- Lower function (inner radius): \( g(x) = x^3 \)
- Interval: \( x = 0 \) to \( x = 1 \)
So the volume becomes:
\[ V = \pi \int_0^1 \left[ (1 + \sin(\pi x))^2 - (x^3)^2 \right] dx = \pi \int_0^1 \left[ (1 + \sin(\pi x))^2 - x^6 \right] dx \]
Option B