Double integral over general region

The blue region is the region type ________

• I

• II

Curves  $x^2=2-y$  and  $x=y$  are shown in this figure. Set up a double integral to find the area of the blue region.

•  $\int_{-2}^1\int_x^{x^2}1\ dydx$  

•  $\int_x^2\int_1^21\ dydx$  

•  $\int_{-2}^1\int_x^{\left(2-x^2\right)}1\ dydx$  

•  $\int_{-2}^0\int_y^{\left(2-y\right)}1\ dxdy$  

 $\int_0^2\int_x^{x^2}x\cdot y^2-x\ dydx$  = __________

Give your answer in decimals. 

The base of a solid is the blue region which is a quarter of the circle  $x^2+y^2=64$  . If the volume of the solid is given by  $\int_0^8\int_0^mf\left(x,y\right)\ dydx$ , find m.

• 8

•  $\sqrt{x^2-64}$  

•  $\sqrt{8-x^2}$  

•  $\sqrt{64-x^2}$  

A cheese with width = 5cm, length = 10cm and height = 5cm is placed at the origin of xyz axes as shown in the figure. Set up a double integral to determine the volume of the cheese.

• $\int_0^{10}\int_0^{\frac{y}{2}}5\ dxdy$

• $\int_0^5\int_0^{2x}1\ dydx$

• $\int_0^{10}\int_0^55\ dxdy$

• $\int_0^5\int_0^52x\ dydx$