Consider the closed region

bounded simultaneously by the paraboloid and plane, jointly denoted

. By the divergence theorem,

And since we have

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have




Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by

, we have

Parameterize

by


which would give a unit normal vector of

. However, the divergence theorem requires that the closed surface

be oriented with outward-pointing normal vectors, which means we should instead use

.
Now,



So, the flux over the paraboloid alone is
Answer:
The shadow is decreasing at the rate of 3.55 inch/min
Step-by-step explanation:
The height of the building = 60ft
The shadow of the building on the level ground is 25ft long
Ѳ is increasing at the rate of 0.24°/min
Using SOHCAHTOA,
Tan Ѳ = opposite/ adjacent
= height of the building / length of the shadow
Tan Ѳ = h/x
X= h/tan Ѳ
Recall that tan Ѳ = sin Ѳ/cos Ѳ
X= h/x (sin Ѳ/cos Ѳ)
Differentiate with respect to t
dx/dt = (-h/sin²Ѳ)dѲ/dt
When x= 25ft
tanѲ = h/x
= 60/25
Ѳ= tan^-1(60/25)
= 67.38°
dѲ/dt= 0.24°/min
Convert the height in ft to inches
1 ft = 12 inches
Therefore, 60ft = 60*12
= 720 inches
Convert degree/min to radian/min
1°= 0.0175radian
Therefore, 0.24° = 0.24 * 0.0175
= 0.0042 radian/min
Recall that
dx/dt = (-h/sin²Ѳ)dѲ/dt
= (-720/sin²(67.38))*0.0042
= (-720/0.8521)*0.0042
-3.55 inch/min
Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min
The scale factor of the dilation is 2
Antonio runs 1,460 miles in one year
Answer:
x^5y+4x+4y+2
Step-by-step explanation:
Fact, ^ is called an exponet
(Just so you know)