Answer
V=96
Work Step by Step
We are given $z=8x+4y+10$ with $-1\leq x \leq1, 0\lt y \lt 3$
Replacing the limits for the region R and $f(x,y)=z$
$V=\int^{3}_{0}\int^{1}_{-1}(8x+4y+10)dxdy$
$V=\int^{3}_{0}[\int^{1}_{-1}(8x+4y+10)dx]dy$
solve the inner integral, treating y as a constant
$\int^{1}_{-1}(8x+4y+10)dx$
integrate $=[4x^{2}+4xy+10x]^{1}_{-1}=4+4y+10-(4-4y-10)=8y+20$
$V=\int^{3}_{0}[\int^{1}_{-1}(8x+4y+10)dx]dy$
$V=\int^{3}_{0}(8y+20)dy$
integrating
$V=4y^{2}+20y|^{3}_{0}=4(3)^{2}+20(3)-(4.0+20.0)=96$
