1. Find 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 $(0,1)$.
Answer
Final Answer: 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 \(\boxed{\frac{5}{8}}\).
Step 1 ：The volume of the solid under the surface \(z=f(x,y)\) and above the region \(D\) in the \(x y\)-plane is given by the double integral \(\int \int_D f(x,y) \, dx \, dy\).
Step 2 ：In this case, \(f(x,y)=1+xy\) and \(D\) is the triangle with vertices \((0,0),(1,1)\), and \((0,1)\).
Step 3 ：We can integrate over \(D\) by integrating \(x\) from \(0\) to \(y\) and \(y\) from \(0\) to \(1\).
Step 4 ：The volume of the solid is calculated to be \(\frac{5}{8}\).
Step 5 ：Final Answer: 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 \(\boxed{\frac{5}{8}}\).
