Step 1 :The first question asks for the volume of the solid that lies under the surface \(z=1+x y\), and above the region \(D\) in the \(x y\)-plane, where \(D\) is a triangle with vertices \((0,0),(1,1)\), and \((0,1)\).
Step 2 :To find the volume of the solid, we can use the double integral of the function \(z=1+x y\) over the region \(D\). The region \(D\) is a triangle, so we can describe it as the set of points \((x, y)\) such that \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\).
Step 3 :The result of the double integral gives the volume of the solid. Therefore, the volume of the solid that lies under the surface \(z=1+x y\), and above the region \(D\) in the \(x y\)-plane, where \(D\) is a triangle with vertices \((0,0),(1,1)\), and \((0,1)\) is 13/24.
Step 4 :Final Answer: The volume of the solid is \(\boxed{\frac{13}{24}}\).