Respuesta :
a) The area within the region is [tex]\frac{16}{3}[/tex].
b) The volume of the solid [tex]S[/tex] is [tex]\pi[/tex] cubic units.
c) The volume of the solid of revolution is [tex]\frac{164}{3}\pi[/tex] cubic units.
How to apply integral formulas to determine areas and volumes
a) The region is bounded between [tex](0,0)[/tex] and [tex](2, 4)[/tex], the area of the region is defined by the following integral formula:
[tex]A = \int\limits^{b}_{a} {[g(x)-f(x)]} \, dx[/tex] (1)
Where:
- [tex]f(x)[/tex] - Lower function
- [tex]g(x)[/tex] - Upper function
- [tex]a[/tex] - Initial value
- [tex]b[/tex] - Final value
If we know that [tex]f(x) = x^{2}[/tex], [tex]g(x) = 4[/tex], [tex]a = 0[/tex] and [tex]b = 2[/tex], then the area within the region is:
[tex]A = \int\limits^{2}_{0} {4-x^{2}} \, dx[/tex]
[tex]A = \frac{16}{3}[/tex]
The area within the region is [tex]\frac{16}{3}[/tex]. [tex]\blacksquare[/tex]
b) The volume of the solid [tex]S[/tex] is determined by the following integral formula:
[tex]V = \frac{\pi}{8} \int\limits^2_0 {x^{2}} \, dy = \frac{\pi}{8} \int\limits^2_0 {x^{2}\cdot (2\cdot x)} \, dx = \frac{\pi}{4}\int\limits^2_0 {x^{3}} \, dx[/tex]
[tex]V = \pi[/tex]
The volume of the solid [tex]S[/tex] is [tex]\pi[/tex] cubic units. [tex]\blacksquare[/tex]
c) The solid of revolution around an axis parallel to the [tex]x[/tex]-axis is described by the following formula:
[tex]V = \pi \int\limits^b_a {[[f(x)-k]^{2}-[g(x)-k]^{2}]} \, dx[/tex] (2)
Where [tex]k[/tex] is the y-coordinate of the axis of revolution.
If we know that [tex]f(x) = x^{2}[/tex], [tex]g(x) = 4[/tex], [tex]k = 7[/tex], [tex]a = 0[/tex] and [tex]b = 2[/tex], then the volume of the solid of revolution is:
[tex]V = \pi\int\limits^2_0 {[(x^{2}-7)^{2}-(4-7)^{2}]} \, dx[/tex]
[tex]V = \frac{164}{3}\pi[/tex]
The volume of the solid of revolution is [tex]\frac{164}{3}\pi[/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
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