Question

Sketch the region bounded by the curves y = x 2 and y = 10 then set up the integral needed to find the volume of the solid generated by revolving this region about the x -axis using the shell method.

Answer 1

Answer:

V=758πvu

Step-by-step explanation:

Data

$y_{1} =x^{2}\\y_{2} =10\\V=?$

The solid of revolution generated by the curves y1 and y2 around  the x-axes must be found

$dV=\pi R^{2}h\\ h=thickness$

$R^{2}=r_{2}^{2}-r_{1}^{2}\\r_{2}=10\\r_{1}=x^{2}\\h=dx\\R^{2}=(10)^{2}-(x^{2})^{2}=100-x^{4}\\dV=\pi (100-x^{4})dx$

to find the cut points we match the curves  y1 and y2

$x^{2} =10\\x=+-\sqrt{10}=+-3.16$

$\int\limits{dV}\,= \pi \int\limits{(100-x^{4})} \, dx\\\int\limits{dV}\,= \pi \int\limits^a_b{(100-x^{4})} \, dx$

where a=3.16 and b=-3.16, but it can also be

$\int\limits {dV} \, dx=2\pi \int\limits^a_0 {(100-x^{4}) } \, dx\\ V=2\pi (100x-\frac{x^{5}}{5});0-3.16\\ V=2\pi((100*3.16)-(-3.16)^{5}/5)\\V=2\pi (316+63.01)=2\pi 379\\V=758\pi vu$