The volume of the solid obtained by rotating the region bounded by the given curves about [tex]y = 1[/tex] is [tex]\frac{11\pi}{30} [/tex] cubic units.
Procedure - Determination of the volume of a solid of revolution
Let [tex]f(x) = 1[/tex], [tex]g(x) = x^{1/2}[/tex] and [tex]h(x) = x^{2}[/tex], both [tex]g(x) [/tex] and [tex]h(x)[/tex] have the following intersection points: [tex](x_{1}, y_{1}) = (0, 0)[/tex] and [tex](x_{2}, y_{2}) = (1,1)[/tex]. This means that the interval of x-coordinates for the solid of revolution is [tex]x \in [0, 1][/tex].
The integral formula for the volume of the solid of revolution ([tex]V[/tex]), in cubic units, is:
[tex]V = \pi \int\limits^{1}_{0} {[h(x)-f(x)]^{2}} \, dx - \pi \int\limits^{1}_{0} {[g(x)-f(x)]^{2}} \, dx [/tex] (1)
After substituting on each function and simplifying terms we get the following integral:
[tex]V = \pi \int\limits^{1}_{0} {(x^{2}-1)^{2}} \, dx - \pi \int\limits^{1}_{0} {(x^{1/2}-1)^{2}} \, dx [/tex]
[tex]V = \pi\left[\int\limits^{1}_{0} {x^{4}} \, dx - 2\int\limits^{1}_{0} {x^{2}} \, dx -\int\limits^{1}_{0} {x} \, dx + 2\int\limits^{1}_{0} {x^{1/2}} \, dx \right][/tex]
And the volume of the solid of revolution is:
[tex]V = \pi \cdot \left(\frac{1}{5}-\frac{2}{3} -\frac{1}{2}+\frac{4}{3} \right)[/tex]
[tex]V = \frac{11\pi}{30} [/tex]
The volume of the solid obtained by rotating the region bounded by the given curves about [tex]y = 1[/tex] is [tex]\frac{11\pi}{30} [/tex] cubic units. [tex]\blacksquare[/tex]
To learn more on solids of revolution, we kindly invite to check this verified question: https://brainly.com/question/338504
