Example 4 find the volume of the solid that lies under the paraboloid z = 6x2 + 6y2, above the xy-plane, and inside the cylinder x2 + y2 = 2x. solution the solid lies above the disk d whose boundary circle has equation x2 + y2 = 2x or, after completing the square, x − 2 + y2 = . in polar coordinates we have x2 + y2 = r2 and x = r cos(θ), so the boundary circle becomes r2 = 2r cos(θ), or r = 2 cos(θ). thus the disk d is given by d = (r, θ) −π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2 cos(θ) and, by this formula, we have v = d (6x2 + 6y2) da = π/2 −π/2 2 cos(θ) 0 6r2 r dr dθ = π/2 −π/2 6 4r4 2 cos(θ) 0 dθ = π/2 0 cos4(θ) dθ = π/2 0 1 + cos(2θ) 2 2 dθ = π/2 0 1 + 2 cos(2θ) + 1 2 (1 + cos(4θ)) dθ

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explanation

the graph has a slope of 1/3 because each time you move up 1, you go to the right 3. so, slope = rise/run = 1/3. we can cross choice a off the list since we want a slope of 3, rather than 1/3.

the first table shown has a slope of -3. this is because the rise is -6 and the run is 2, so slope = rise/run = -6/2 = -3. note how x increases by 2 each time (run) and y drops by 6 each time (rise). choice b can be eliminated.

the second table has a slope of 3. each time x goes up by 1, the y values increase by 3. slope = rise/run = 3/1 = 3. alternatively you can use the slope formula to prove that the slope is 3 here. use any two rows of values to generate the two points. choice c is one of the answers

the third table also has a slope of 3. let's use the slope formula to prove as such. i'm going to use the first two rows of the table. so i'll use the two points (2,1) and (5,10)

m = slope

m = (y2-y1)/(x2-x1)

m = (10-1)/(5-2)

m = 9/3

m = 3 so we get a slope of 3

choice d is the other answer

i think the answer is true

step-by-step explanation:

the weight of the object inside the box!

40 i don’t know lololol