Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. \( y=x^{4}, \quad y=\sqrtx, about the x-axis V=∫ a

b
​
A(x)dx=∫ 0
1
​
π([x)dx

Answer 1

To find the volume of a solid obtained by rotating the region bounded by the curves \(y=x^{4}\) and \(y=\sqrt{x}\) about the x-axis, we will use the method of washers. The volume V of a solid of revolution can be found by the integral \(V=\int_{a}^{b} A(x)dx\), where A(x) represents the cross-sectional area of the solid at any point x in the interval [a, b].

The x-values where the two functions intersect are the limits of the integral, which in this case are a = 0 and b = 1.

The cross-sectional area A(x) of the solid can be found by subtracting the area of the smaller disk (the one closer to the x-axis) from the area of the larger disk. Given the functions are \(y=g(x)=\sqrt{x}\) for the outer radius (further from x axis) and \(y=f(x)=x^{4}\) for the inner radius (closer to x axis), and the area of a disk with radius \(r\) being modelled as \(\pi r^{2}\), we can write the area function as follows: A(x) = \(\pi\left[g(x)\right]^2 - \pi\left[f(x)\right]^2\).

Substituting the functions into A(x) gives: A(x) =
`\(\pi\left[√(x)\right]^2 - \(\pi\left[x^4\right]^2\).`

Substituting the limits a and b and the area function, the volume integral becomes:

\[V=\int_{a}^{b} A(x)dx = \int_{0}^{1} \pi(\sqrt{x})^2 - \pi(x^4)^2 dx\].

This is a definite integral and can be evaluated using techniques of integral calculus.

Carrying out the integration yields the volume of the solid:


`\[V = 1.2217304763960306\]`

So, the volume of the solid obtained by rotating the region bounded by the curves
`\(y=x^(4)\) and \(y=√(x)\)` about the x-axis is approximately 1.222 cubic units.