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Sagot :
Using integrals, the volume generated by rotating the given region about the specified line is of 0.8π cubic units.
How to use integrals to find the volume of a curve?
Rotating a function f(x) around a function g(x), the volume is given by the following integral:
[tex]V = \pi \int_{a}^{b}[f(x) - g(x)]^2 dx[/tex]
In which a and b are the intersection points of the curve.
For this problem, from the graph, we have that:
- [tex]f(x) = 2\sqrt[4]{x}[/tex].
- g(x) = 2x.
- a = 0.
- b = 1.
Hence:
[tex]V = \pi \int_{0}^{1}[2\sqrt[4]{x} - 2x]^2 dx[/tex]
[tex]V = \pi [\int_{0}^{1} 4\sqrt{x} - 8x\sqrt{x} + 4x^2 dx][/tex]
[tex]V = \pi \left[\frac{8}{3}x^{\frac{3}{2}} - \frac{16}{5}x^{\frac{5}{2}} + \frac{4}{3}x^3}\right]_{x=0}^{x=1}[/tex]
[tex]V = \pi \left[\frac{8}{3} - \frac{16}{5} + \frac{4}{3}\right][/tex]
[tex]V = \pi \left[4 - \frac{16}{5}\right][/tex]
[tex]V = \frac{4\pi}{5}[/tex]
V = 0.8π cubic units.
More can be learned about integrals and volumes at https://brainly.com/question/18371476
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