Answer
$\frac{3}{4}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region D in the xy-plane can be defined as:
$$Volume=\int\int_Df(x,y)dA$$
We can express the region D as follows:
$$D=\{(x,y)|0\leq x\leq 1,x^2\leq y\leq \sqrt x\}$$
Therefore,
$$Volume=\int\int_Df(x,y)dA
\\=\int_0^1\int_{x^2}^\sqrt{x}(3x+2y)dydx
\\=\int_0^1[3xy+y^2]_{y=x^2}^{y=\sqrt{x}}dx
\\=\int_0^1(-x^4-3x^3+3x^\frac{3}{2}+x)dx
\\=[-\frac{1}{5}x^5-\frac{3}{4}x^4+\frac{6}{5}x^\frac{5}{2}+\frac{1}{2}x^2]_0^1
\\=\frac{3}{4}$$
