In Exercises $55-58$, find the signed volume between the graph of the given function and the $xy$-plane over the specified rectangle in the$xy$-plane.

$f\left(x,y\right)=3x-2y^5+1$,

where$R=\left\{\left(x,y\right)|-4\le x\le 6and0\le y\le 7\right\}$

We have given the following function :-

$f\left(x,y\right)=3x-2y^5+1$

where localid="1650627302142" $R=\left\{\left(x,y\right)|-4\le x\le 6and0\le y\le 7\right\}$

We have to find the volume between the graph of the function$f\left(x,y\right)$and the specified rectangle in the$xy-$plane.

We have given the following function :-

$f\left(x,y\right)=3x-2y^5+1$

where $R=\left\{\left(x,y\right)|-4\le x\le 6and0\le y\le 7\right\}$

We can see that localid="1650639473202" role="math" $f\left(x,y\right)\ge 0$on the rectangle $R_1=\left\{\left(x,y\right)|0\le x\le 6and0\le y\le 7\right\}$and localid="1650639486793" $f\left(x,y\right)\ge 0$on the rectangle $R_2=\left\{\left(x,y\right)|-4\le x\le 0and0\le y\le 7\right\}$

Therefore the required volume is given by the following difference :-

localid="1650639522202" role="math" $\int {\int }_R\left|3x-2y^5+1\right|dA=\int {\int }_{R_1}\left(3x-2y^5+1\right)dA+\int {\int }_{R_2}\left(3x-2y^5+1\right)dA$

Then by using Fubini's theorem, we have :-

localid="1650637790591" $\int {\int }_{R_1}\left(3x-2y^5+1\right)dA={\int }_0^7{\int }_0^6\left(3x-2y^5+1\right)dxdy$and $\int {\int }_{R_2}\left(3x-2y^5+1\right)dA={\int }_0^7{\int }_{-4}^0\left(3x-2y^5+1\right)dxdy$

Then :-

$$\begin{gathered} \int {\int }_{R_1}\left(3x-2y^5+1\right)dA+\int {\int }_{R_2}\left(3x-2y^5+1\right)dA \\ ={\int }_0^7{\int }_0^6\left(3x-2y^5+1\right)dxdy+{\int }_0^7{\int }_{-4}^0\left(3x-2y^5+1\right)dxdy \\ ={\int }_0^7\left({\int }_0^6\left(3x-2y^5+1\right)dx\right)dy+{\int }_0^7\left({\int }_{-4}^0\left(3x-2y^5+1\right)dx\right)dy \\ ={\int }_0^7{{\left[\frac{3x^2}{2}-2y^{5}x+x\right]}^6}_{0}dy+{\int }_0^7{{\left[\frac{3x^2}{2}-2y^{5}x+x\right]}^0}_{-4}dy \\ ={\int }_0^7\left[\left(\frac{3\times 6\times 6}{2}-12y^5+6\right)-0\right]dy+{\int }_0^7\left[0-\left(\frac{3\times -4\times -4}{2}+8y^5-4\right)\right]dy \\ ={\int }_0^7\left(54-12y^5+6\right)dy+{\int }_0^7\left(-24-8y^5+4\right)dy \\ ={\int }_0^7\left(60-12y^5\right)dy+{\int }_0^7\left(-20-8y^5\right)dy \\ ={{\left[60y-\frac{12y^6}{6}\right]}^7}_0+{{\left[-20y-\frac{8y^6}{6}\right]}^7}_0 \\ =\left[\left(420-2\times 7^6\right)-0\right]+\left[-140-\frac{4\times 7^6}{3}\right] \\ =\left[420-235298\right]+\left[-140-\frac{470596}{3}\right] \\ =-234878+\left[\frac{-420-470596}{3}\right] \\ =-234878+\left[-\frac{471016}{3}\right] \\ =-234878-\frac{471016}{3} \\ =\frac{-704634-471016}{3} \\ =\frac{-1175650}{3} \end{gathered}$$