Answer:
Step-by-step explanation:
If you graph there would be two different regions. The first one would be
[tex]y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\[/tex]
And the second one would be
[tex]y = 2-x \,\,\,\,\,, 1 \leq x \leq 2[/tex].
If you rotate the first region around the "y" axis you get that
[tex]{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }[/tex]
And if you rotate the second region around the "y" axis you get that
[tex]{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }[/tex]
And the sum would be 2.51+4.188 = 6.698
If you revolve just the outer curve you get
If you rotate the first region around the x axis you get that
[tex]{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }[/tex]
And if you rotate the second region around the x axis you get that
[tex]{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }[/tex]
And the sum would be 1.5708+1.0472 = 2.618
