Use a triple integral to find the volume of the solid bounded by the graphs...

Step 1
Given:
The solid bounded by the graphs of the equations:
z=2−y and ${\displaystyle z=4-y^2}$
x=0 and x=3
y=0
We have to find the volume of the solid bounded by the given graphs of the equations.
Step 2
We know that,
The limits for z is:
${\displaystyle z=2-y\to z=4-y^2}$
The limit for x is:
x=0 to x=3
Now we have to find the limit of y:
We have,
${\displaystyle z=2-y\to z=4-y^2}$
${\displaystyle \Rightarrow 2-y=4-y^2}$
${\displaystyle \Rightarrow y^2-y-2=0}$
${\displaystyle \Rightarrow y^2-2y+y-2}$
${\displaystyle \Rightarrow y(y-2)+1(y-2)=0}$
${\displaystyle \Rightarrow (y-2)(y+1)=0}$
${\displaystyle \Rightarrow (y-2)=0\text{or}(y+1)=0}$
y=2 or y =-1
y=-1 is not possible ${\displaystyle [y=0]}$
Hence we get the limit for y is:
y=0 to y=2
Step 3
${\displaystyle V={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}{\int }_{z=2-y}^{z=4-y^2}dzdydx}$
${\displaystyle ={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}{\left[z\right]}_{2-y}^{4-y^2}dydx}$
${\displaystyle ={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}[4-y^2-2+y]dydx}$
${\displaystyle ={\int }_{x=0}^{x=3}{\int }_{y=0}^{y=2}[2-y^2+y]dydx}$
${\displaystyle ={\int }_{x=0}^{x=3}{[2y-\frac{y^3}{3}+\frac{y^2}{2}]}_0^{2}dx}$
${\displaystyle ={\int }_{x=0}^{x=3}[4-\frac{8}{3}+2]dx}$
${\displaystyle =\frac{10}{3}{\int }_{x=0}^{x=3}dx}$
${\displaystyle =\frac{10}{3}{\left[x\right]}_0^3}$