Another Volume Problem...???

Question

The 1st quadrant area bound by y=x^2 and y=x+2 is revolved around the vertical axis. What is the volume generated?

Dr. D, shouldn't dV = pi*[y-(y-2)]^2dy ???

Dr. D, I stand corrected. Thanks, sweet2water.

Answer 1

Since you're revolving around the y axis, you need to integrate
π*x^2*dy between y = 0 and y = 4.
The curves intersect at (2,4).

Integral of π * [y - (y-2)^2]*dy
= π *[y^2/2 - (y-2)^3 / 3]
= π * [8 - (8/3 + 8/3)]
= 8π/3

Answer 2

You can solve this problem by first finding the point of intersection between the two curves to find your limits. Area will be bound between the points x=0 and x=2. The formula to find volume of revolution is the integral from a to b of 2πx times (x+2-x^2)dx. The term in parentheses represents the height of the cross section at any point x (basically the difference beween the two functions, being careful to put the more positive function first). When you evaluate your integral, you get 2π[(x^3)/3+x^2-(x^4)/4] evaluated between 0 and 2. Your final answer is 16π/3.

Edit: The method of revolution mentioned by Dr D does not work. The problem states that the area has to be first quadrant only. Yet the function y = x+2 enters the second quadrant below y = 2. Therefore your limits will only work between y = 2 and y = 4. Between y = 0 and y = 2, you have to use the function x = 0 as your left bound. I find my method simpler and accurate.

Answer 3

do exactly dimensional conversion by technique of dividing 40 5 g by technique of two.sixty seven g/ml the gram contraptions cancel. you're left with sixteen.80 5 mL this is .001 L multiply those 2 to get the quantity of liters, 0.016L.

Answer 4

Inday, the good doctor is correct: it's the difference of the squares, not the square of the difference.