Can you please check to see if these answers are correct? I have provided my work as well! That you! This is much appreciated!!
1. What is the volume of the solid that is generated by revolving the region bounded by the curves x = 3y^2-2 and x = y^2 and y = 0 about the x axis?
Shell Method: h = y
r = y^2-(3y^2-2)
My Work: 2pi∫ [0,1] (y)(y^2-3y^2+2)dy
2pi((-2(1)^4/4)+2(1)^2/2)
Answer = 3pi
2. What is the volume of the solid that is generated by revolving the region bounded by the curves x = 3y^2-2 and x = y^2 and y = 0 about the vertical line x = 1?
Washer Method: R = 1 - (3y^2-2)
r = 1-y^2
My Work: pi∫ [0,1] (-3y^2+3)^2-(-1+y^2)^2 dy
pi (8(1)^5/5 - 16(1)^3/3 + 8(1))
Answer = 64pi/15
#1. Using horizontal shells of thickness dy,
r = y
h = y^2 - (3y^2-2) = 2-2y^2
v = ∫[0,1] 2πrh dy
= 2π∫[0,1] y(2-2y^2) dy = π
You got r and h reversed, but the product was luckily the same. Unfortunately, you added instead of subtracting.
Using discs of radius y and thickness dx,
v = ∫π(R^2-r^2) dx
v = ∫[-2,0] π(x+2)/3 dx + ∫[0,1] π((x+2)/3 - x) dx = 2π/3 + π/3 = π
#2. looks good.
Just as a check, using shells of thickness dx,
v = ∫2πrh dx
= ∫[-2,0] 2π(1-x)√((x+2)/3) dx
+ ∫[0,1] 2π(1-x)(√((x+2)/3)-(√x)) dx
= 8π/5 √6 + (64/15 π - 8π/5 √6)
= 64/15 π
Thank you Steve!
To check if the provided answers are correct, let's go through the steps of each method and calculate the volume.
1. Shell Method:
The formula for the volume using the shell method is V = 2π∫[a, b] (r)(h)dx.
In this case, the height (h) is given as y. The radius (r) is calculated as y^2 - (3y^2 - 2), which simplifies to 2 - 2y^2.
Using the formula and limits of integration, the integral becomes:
V = 2π∫[0, 1] (2 - 2y^2)(y)dy
Evaluating the integral gives:
V = 2π ∫[0, 1] (2y - 2y^3)dy
V = 2π [y^2 - (1/2)y^4] evaluated from 0 to 1
Plugging in the limits and simplifying:
V = 2π [(1) - (1/2)]
V = π(1 - 1/2)
V = π(1/2)
V = π/2
The provided answer of 3π is incorrect. The correct answer is π/2.
2. Washer Method:
The formula for the volume using the washer method is V = π∫[a, b] [(R)^2 - (r)^2]dx.
In this case, the outer radius (R) is 1 - (3y^2 - 2), and the inner radius (r) is 1 - y^2.
Using the formula and limits of integration, the integral becomes:
V = π∫[0, 1] [(1 - (3y^2 - 2))^2 - (1 - y^2)^2]dy
Expanding and simplifying the expression inside the integral:
V = π∫[0, 1] [(8y^4 - 12y^2 + 5)(5y^2 - 2)]dy
Evaluating the integral gives:
V = π [(8/5)y^7 - 12/3)y^5 + (5/3)y^4 - (2/5)y^2] evaluated from 0 to 1
Plugging in the limits and simplifying:
V = π [ (8/5) - (12/3) + (5/3) - (2/5) ]
V = π [ (8/5) - (20/5) + (25/15) - (6/15) ]
V = π [ (8 - 20 + 25 - 6) / 15 ]
V = π [ 7/15 ]
The provided answer of 64π/15 is incorrect. The correct answer is 7π/15.
Remember to double-check the calculations and make sure to use the correct formulas and limits of integration.