Answer
a. $60$
b. $36$
Work Step by Step
a. Step 1. Draw a diagram as shown in the figure. The limits of integration with respect to $x$ can be found from $0$ to $2$.
Step 2. The cross section is a rectangle with one side length of $u=6-3x$. Assuming the other length is $v$, we have $dV=uv\ dx=(6-3x)v\ dx$
Step 3. For $v=10$, we have $V=\int_0^2 10(6-3x)dx=(60x-15x^2)|_0^2=60(2)-15(2)^2=60$
b. With a perimeter of $20$, we have $2(u+v)=20$ and $v=10-u=4+3x$. Thus:
$V=\int_0^2 (6-3x)(4+3x)dx=\int_0^2 (24+6x-9x^2)dx=(24x+3x^2-3x^3)|_0^2=24(2)+3(2)^2-3(2)^3=36$
