Question

g) Find the volume of the solid generated by rotating the
parabola y2 = 4ax about the x-axis between x=0 and
X = a.​

Answer 1

Answer:

xpress y as a function of x for the first quadrant: y=2ax−−√. Use the restriction that a>0 (for negative values of a, you would get the same volume because the solid is symmetric with respect to the axis of revolution) and shift the curve along the x-axis a units to the right: y=2a(x−a)−−−−−−−√. And then you're going to use shell integration with the bounds of a and 2a

:

V=2π∫2aax⋅2a(x−a)−−−−−−−√dx=4π∫2aaxax−a2−−−−−−√dx.

You can use the substitution u=ax−a2

to solve that integral:

4πa2∫2aa(ax−a2+a2)ax−a2−−−−−−√ddx(ax−a2)dx=4πa2∫a20(u+a2)u−−√du=4πa2∫a20(u32+a2u12)du.

From this point on, the integration is trivial.

Step-by-step explanation: