IAT 2030
HHW \#8
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 $(0,1)$.
2. Use polar coordinates to evaluate the double integral,
\[
\iint_{D} x d A
\]
where $D$ is the region in the first quadrant that lies between the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}=2 x$
3. Evaluate the integrals by changing order of integration.
\[
\int_{0}^{1} \int_{\arcsin y}^{\pi / 2} e^{\cos x} d x d y
\]
4. Use polar coordinates to find the volume of the solid that is:
inside the sphere $x^{2}+y^{2}+z^{2}=16$ and outside the cylinder $x^{2}+y^{2}=4$.

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 \(\boxed{\frac{13}{24}}\).