Atank in the shape of a right circular cone has height 12 feet and base radius 4 feet. the tank is inverted, with its vertex pointing down and base at the top. the tank contains a liquid with weight density 63 pounds per cubic foot, but is filled to a depth of 8 feet only. set up, but do not evaluate, an integral for the work (in foot-pounds) required to pump all the liquid present to a height one foot over the top of the tank.

V = int(π(y/3)^2, 0, 8)

(Definite integration of π(y/3)^2 with lower boundary 0 and upper boundary 8)

Step-by-step explanation:

Set up cartesian axis (x and y) to the system.

Let y axis as the line of the centre of the cone, passing through its vertex and the centre of it's circular base. The x axis could be the 90 degree line to the y axis that passes the vertex. So the origin (0,0) is at the vertex.

I'm this setup , looking it as if we are looking it in 2 dimension, we'll see that there is a signature straight line on the x-y plane, which this line will form the cone as it revolute around the y-axis

Find the equation of the line:

Using height 12 and radius base 4, we can get the slope of the line

m = 12/4 = 3

It passes through origin, so the y-intercept is 0

Hence, y = 3x

Since the volume revolves around y-axis, we use the equation volume of revolution around y-axis

V = int(πx^2,a,b)

(Definite integration of πx^2 with lower boundary a and upper boundary b

Since y=3x

x = y/3

For this que

V = int(π(y/3)^2, 0, 8)

(Definite integration of π(y/3)^2 with lower boundary 0 and upper boundary 8)

step-by-step explanation: step 3 should really go to step seven, but anyway never mind.

1 ∠oln ≅ ∠lno

spinning a 1, 2, or 3. option d.

step-by-step explanation: when spinning a number graph up to 4, and aiming for any number less than 3, that automatically eliminates the number 4. now you have two numbers to spin on that are less than 3, but the number 3 is included in this question because it's no greater of an number than itself. i hope this you : )