Sketch the region bounded by the curves y=9x3,y=9 and x=0 then find the volume of the solid generated by revolving this region about the x-axis.

Volumes of revolution are so fun
lets say you have 2 functions f(x) and g(x) where they intersect at c and d where c<d, and f(x) is on top of g(x) (like for a value r where c<r<d, f(r)>g(r)), the volume is


ok, so find the intersection points of and y=9

9=9x^3
1=x^3
at x=1

so then the area will be from x=0 to x=1
which ones's higher?
try x=0.5
at x=0.5, 9(0.5)^3<9
so the y=9 is on top
therfor

limits are from 0 to 1
higher one is y=9
bottom function is y=9x^3
so the volume is


we can factor out the 81

factor out the constant

integrate





the area is square units

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !