Answer:
[tex] V = \pi [4\sqrt{2} -\frac{8\sqrt{2}}{3} +1.131] =9.478[/tex]
Step-by-step explanation:
For this case we have the following function [tex] y = 2-x^2[/tex] bounded by [tex] x,y =0[/tex]. The area of interest is on the figure attached.
For this case we can calculate the volume using the method of rings. First we can calculate the area like this:
[tex] A(x) = \pi r^2= \pi (2-x^2)^2 = \pi(4-4x^2 + x^4)[/tex]
And the volume can be founded with the following formula:
[tex] V= \int_{a}^b A(x) dx [/tex]
For this case we need to find the intersection point with the x axis, and we can do this:
[tex] 2-x^2 = 0[/tex]
[tex] x= \pm \sqrt{2}[/tex], but for our case would be just the positive value, So then we can find the volume like this:
[tex] V= \pi \int_{0}^{\sqrt{2}} 4 -4x^2 +x^4 dx[/tex]
And if we do the integral we got this:
[tex] V = \pi [4x -\frac{4}{3}x^3 +\frac{x^5}{5} \Big|_{0}^{\sqrt{2}}][/tex]
And when we apply the fundamental theorem of calculus we got:
[tex] V = \pi [(4\sqrt{2} -\frac{4}{3} (\sqrt{2})^3 +\frac{(\sqrt{2})^5}{5}) -(0)][/tex]
[tex] V = \pi [4\sqrt{2} -\frac{8\sqrt{2}}{3} +1.131] =9.478[/tex]
