Find the area of the surface given by $z=$ $f(x, y)$ over the region $R$. (Hint: Some of the integrals are simpler in polar coordinates.) \

$$f(x, y)=9-x^2$$

$R$ : square with vertices $(0,0),(3,0),(0,3),(3,3)$

Find the area of the surface given by $z=$ $f(x, y)$ over the region $R$. (Hint: Some of the integrals are simpler in polar coordinates.)
$f(x, y)=2 x+2 y$ $R$ : triangle with vertices $(0,0),(2,0),(0,2)$

In the following exercise, find the area of the surface given by z=f(x, y) over the region R. (Hint: Some of the integrals are simpler in polar coordinates.)

$$f(x, y)=2 x+2 y$$

$R$ : triangle with vertices $(0,0),(2,0),(0,2)$

Find the area of the surface given by $z=$ $f(x, y)$ over the region $R$. (Hint: Some of the integrals are simpler in polar coordinates.)
$f(x, y)=10+2 x-3 y$

$$R=\left\{(x, y): x^2+y^2 \leq 9\right\}$$

Find the area of the surface given by z = f(x, y) over the region R. (Hint:Some of the integrals are simpler in polar coordinates.) f(x, y) = 7 + 2x + 2y, R = {(x, y): x² + y² ≤ 4}

Find the area of the surface. -The portion of the paraboloid z = 16 - x² - y² in the first octant

Find the area of the surface given by z = f(x, y) over the region R. (Hint:Some of the integrals are simpler in polar coordinates.) f(x, y) = √(a² - x² - y²) R = {(x, y): x² + y² ≤ a²}

Find the area of the surface given by z=f(x, y) that lies above the region R. $f(x, y)=4+5 x+6 y, \ R=\left\{(x, y): x^{2}+y^{2} \leq 4\right\}$

Find the area of the surface given by z = f(x, y) over the region R. (Hint:Some of the integrals are simpler in polar coordinates.) f(x, y) = 2x + 2y R: triangle with vertices (0, 0), (4, 0), (0, 4)

Find the area of the surface given by z = f(x, y) over the region R. f(x, y) = y² R: square with vertices (0, 0), (3, 0), (0, 3), (3, 3)

Find the area of the surface given by z = f(x, y) over the region R. (Hint:Some of the integrals are simpler in polar coordinates.) f(x, y) = √x²+y², R = {(x, y): 0 ≤ f(x, y) ≤ 1}

In the following exercise, use a double integral to find the area of the shaded region.

given the function $r=2+\sin \theta$

Use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. z = ln(x² + y²), z = 0, x² + y² ≥ 1, x² + y² ≤ 4

Use a double integral in polar coordinates to find the volume of the solid bounded by the graphs of the equations. z = √25-x²-y², z = 0, x² + y² = 16