Question

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.

Answer 1

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
$\pi \int\limits^d_c {f(x)^2-g(x)^2} \, dx$


ok, so find the intersection points of $y=9x^3$ 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
$\pi \int\limits^1_0 {(9)^2-(9x^3)^2} \, dx =$
$\pi \int\limits^1_0 {81-81x^6} \, dx =$
we can factor out the 81
$\pi \int\limits^1_0 {81(1-x^6)} \, dx =$
factor out the constant
$81 \pi \int\limits^1_0 {1-x^6} \, dx =$
integrate
$81 \pi [x-\frac{1}{7}x^7]^1_0=$
$81 \pi [(1-\frac{1}{7}(1)^7)-(0-\frac{1}{7}(0)^7)]=$
$81 \pi (1-\frac{1}{7}-0)=$
$81 \pi -\frac{81}{7}$

the area is $81 \pi-\frac{81}{7}$ square units