Annette Arroyo

2021-01-31

Use a triple integral to find the volume of the solid bounded by the graphs of the equations. $z=2-y,z=4-{y}^{2},x=0,x=3,y=0$

timbalemX

Skilled2021-02-01Added 108 answers

Step 1

Given:

The solid bounded by the graphs of the equations:

z=2−y and$z=4-{y}^{2}$

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:

$z=2-y\to z=4-{y}^{2}$

The limit for x is:

x=0 to x=3

Now we have to find the limit of y:

We have,

$z=2-y\to z=4-{y}^{2}$

$\Rightarrow 2-y=4-{y}^{2}$

$\Rightarrow {y}^{2}-y-2=0$

$\Rightarrow {y}^{2}-2y+y-2$

$\Rightarrow y(y-2)+1(y-2)=0$

$\Rightarrow (y-2)(y+1)=0$

$\Rightarrow (y-2)=0{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}(y+1)=0$

y=2 or y =-1

y=-1 is not possible$[y=0]$

Hence we get the limit for y is:

y=0 to y=2

Step 3

$V={\int}_{x=0}^{x=3}{\int}_{y=0}^{y=2}{\int}_{z=2-y}^{z=4-{y}^{2}}dzdydx$

$={\int}_{x=0}^{x=3}{\int}_{y=0}^{y=2}{\left[z\right]}_{2-y}^{4-{y}^{2}}dydx$

$={\int}_{x=0}^{x=3}{\int}_{y=0}^{y=2}[4-{y}^{2}-2+y]dydx$

$={\int}_{x=0}^{x=3}{\int}_{y=0}^{y=2}[2-{y}^{2}+y]dydx$

$={\int}_{x=0}^{x=3}{[2y-\frac{{y}^{3}}{3}+\frac{{y}^{2}}{2}]}_{0}^{2}dx$

$={\int}_{x=0}^{x=3}[4-\frac{8}{3}+2]dx$

$=\frac{10}{3}{\int}_{x=0}^{x=3}dx$

$=\frac{10}{3}{\left[x\right]}_{0}^{3}$

Given:

The solid bounded by the graphs of the equations:

z=2−y and

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:

The limit for x is:

x=0 to x=3

Now we have to find the limit of y:

We have,

y=2 or y =-1

y=-1 is not possible

Hence we get the limit for y is:

y=0 to y=2

Step 3