The volume of the solid generated by rotating the region bounded by the curves y = 0, y = x² + x, x = 2 and x = 0 about the y-axis is 40π/3 cubic units.
To find the volume of the solid generated, we use the shell method since the cross-sections are parallel to the axis of rotation.
So, [tex]V = 2\pi\int\limits^a_b {xy} \, dx[/tex]
Now given that the region bounded by the curves y = 0, y = x² + x, x = 2 and x = 0 about the y-axis.
So, we integrate from x = 0 to x = 2.
[tex]V = 2\pi\int\limits^a_b {xy} \, dx\\= 2\pi\int\limits^2_0 {x(x^{2} + x)} \, dx\\= 2\pi\int\limits^2_0 {(x^{3} + x^{2} )} \, dx\\= 2\pi\int\limits^2_0 {x^{3} + 2\pi\int\limits^2_0 x^{2} } \, dx\\= 2\pi[\frac{x^{4} }{4}]_{0}^{2} + 2\pi[\frac{x^{3} }{3}]_{0}^{2} \\= 2\pi[\frac{2^{4} - 0^{4} }{4}}]+ 2\pi[\frac{2^{3} - 0^{3} }{3}]\\= 2\pi[\frac{16 - 0}{4}}]+ 2\pi[\frac{8 - 0}{3}]\\= 2\pi[\frac{16}{4}}]+ 2\pi[\frac{8}{3}]\\= 2\pi (4)+ [\frac{16\pi}{3}]\\= 8\pi + [\frac{16\pi}{3}]\\[/tex]
[tex]= \frac{24\pi + 16\pi}{3} \\= \frac{40\pi}{3}[/tex]
So, the volume of the solid generated by rotating the region bounded by the curves y = 0, y = x² + x, x = 2 and x = 0 about the y-axis is 40π/3 cubic units.
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