Thread: Find Volume of Solid S

Find the volume V of the described solid S. A pyramid with height 3h and a rectangular base with dimensions 3b and 6b.

So I found the following: 3b[1-y/3h) and 6b(1-y/3h) --> Which I belive is where the error is coming from because I cant seem to get the correct answer.

Integral from a=0,b=h of 18h^2(1-y/3h)^2 dy

= from 0 to h of y - 2/3h(1/2)y^2 + 1/9h^2(1/3)y^3

= 38/3b^2h

Thanks in advance.

could you explain what is y?
The volume of Solid S should be 18h*b^2,
The volume of a pyramid is 1/3 * Base Area * Height

Originally Posted by piscoau

could you explain what is y?
The volume of Solid S should be 18h*b^2,
The volume of a pyramid is 1/3 * Base Area * Height

I'm using intergral calculus, or attempting to, to find the volume of the described solid. Using the shells method. So y is the variable in this case denoting the height of the solid if rotated about the y-axis.

I think I figured it out, the limits of integration were were, it should be from a=0 to b=3h