Volume of solid?

Question

I need to find the volume of the solid obtained by rotating the region given by the given curves about the y-axis bounded by

y=36x-6x^2 & y=0

so I have plotted the graph and found the points of intersection to be at 0 and 6

I found the expression of the indefinite integral:

-2x^3 +18x^2 + C

Then I found the parameters I will use to find the volume of my solid which I found by finding the maximum value of y=36x-6x^2 by taking the derivative and setting it to zero. This gave me a value of 3.

So I integrate the function from [0,3] but this only gives me the cross-sectional area.

How do I use the washer method to finish this question? or... did I approach it wrong?

Answer 1

Since you're rotating about the y axis, it screws things up a bit. First draw a sketch so you'll see what's happening. On the sketch, draw a thin vertical rectangle x units fromt he y axis with width, dx and height, y.

This elemental area, dA = y*dx
We need to rotate this about the y axis.
Clearly you'll get a cylinder with radius x
dV = volume of elemental cylinder = 2πx*dA
= 2πx*y*dx
= 2π*x*(36x - 6x^2)*dx
= 2π*(36x^2 - 6x^3)*dx

To find the total volume, you need to sum up all these elemental cylinders, or integrate from x = 0 to 6
V = 2π [12x^3 - 1.5*x^4] |x = 0 to 6
= 1296π

Answer 2

personally I would have gone at it by cylinders. if you try and do by washers the inner and outer diameter are functions of x. by using cylinders only one value has to be calculated.
so if the the volume is the sum of cylinders of r=x from 0~6 and the hight is y=36x-6x^2 then the volume is the integral from 0~6 of 2 pi x f(x) dx