- Thread starter
- #1

The object isn't a prism or pyramid so I am not sure what to do.

View attachment c05.pdf

- Thread starter dwsmith
- Start date

- Thread starter
- #1

The object isn't a prism or pyramid so I am not sure what to do.

View attachment c05.pdf

- Admin
- #2

\(\displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx=\frac{5}{4}\)

- Thread starter
- #3

How did you derive this formula? Is the 1 the volume of the cube or is that part of the triangular shape?

\(\displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx=\frac{5}{4}\)

- Admin
- #4

\(\displaystyle dV=\frac{1}{2}bh\,dx\)

where the base is a constant 1 and the height is $x$, hence:

\(\displaystyle dV=\frac{1}{2}x\,dx\)

and so summing the slices (and adding in the cubical portion), we find:

\(\displaystyle V=1+\frac{1}{2}\int_0^1 x\,dx\)