91Ó°ÊÓ

We have given the following function :-

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

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

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

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

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

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

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

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

Then :-

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