Calculus 3: Surface area

Find the surface area of the part of the surface $z = x^2 + 2y$ that lies above the triangular region with vertices (0, 0), (1, 0), and (1, 1).

Find the surface area of the portion $z = x^2 + y^2$ below the plane $z = 9$.

Find the surface area of the curve $z = \sqrt{4 - x^2}$ above $R$ which is the rectangle where $x$ is between $0$ and $1$ and $y$ is between $0$ and $4$.

Find the surface area of the part of the surface $Z = x^2 + 2y$ that lies above the triangular region with vertices $(0, 0)$, $(1, 0)$, and $(1, 1)$.

Find the surface area of $Z = \sqrt{4 - x^2}$ which is above the region $R$, a rectangle defined by $x \in [0, 1]$ and $y \in [0, 4]$.

Compute the surface area of a surface given its parametric description. Use the formula: $\ iint_{\text{Region}} \| \mathbf{r}_u \times \mathbf{r}_v \| \, du \, dv$ where $\mathbf{r}_u$ and $\mathbf{r}_v$ are the partial derivatives of the position vector with respect to the parameters $u$ and $v$.

Find the surface area of the portion of the plane described by $z = -x$ that is inside the cylinder $x^2 + y^2 = 4$.