conman
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Can some one help with following questions:
4 Units Volume (1 Viewer)
- Thread starter conman
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1) Latus rectum is an interval of the line x = 2 for that equation. When rotated, the radius of the disc is 2 - x (draw a diagram to deduce this) and its thickness is δy, therefore:
δV = π(2 - x)²δy
The equation is y² = 8x
.: x = y²/8
V = π∫(2 - y²/8)²dy with limits from -4 to 4 (the y values where the latus rectum cuts the curve). Expand and evaluate the integral and you should get 256π/15
2) Draw a diagram and the radius of the disc should be 6 - x. It is very similar to the previous one in terms of steps involved. Just substitute y and y value limits which should give you a volume of 128π/15
3) This slice gives an annulus, so when drawing the diagram you'll notice symmetry in the parabola. The radius of the hollow part (the smaller radius) is x, and using the symmetry at x = 1, the larger radius is x + 2(1 - x) which is 2 - x. Threfore, the total area A = π[(2 - x)² - x²] = π(4x + 4) = 4π(x + 1)
The equation is y = 2x - x², so
- y = x² - 2x
1 - y = x² - 2x + 1
1 - y = (x - 1)²
.: x - 1 = √(1 - y)
Sub that into A such that A = 4π√(1 - y)
To find volume, the small change in volume δV = 4π√(1 - y)δy
V = 4π∫√(1 - y) dy with limits 0 to 1 (the limiting y values of the volume)
Evaluate the integral and you should get 8π/3
For the last one, check if you got the equations in the question correct, because those two curves don't even intersect.
δV = π(2 - x)²δy
The equation is y² = 8x
.: x = y²/8
V = π∫(2 - y²/8)²dy with limits from -4 to 4 (the y values where the latus rectum cuts the curve). Expand and evaluate the integral and you should get 256π/15
2) Draw a diagram and the radius of the disc should be 6 - x. It is very similar to the previous one in terms of steps involved. Just substitute y and y value limits which should give you a volume of 128π/15
3) This slice gives an annulus, so when drawing the diagram you'll notice symmetry in the parabola. The radius of the hollow part (the smaller radius) is x, and using the symmetry at x = 1, the larger radius is x + 2(1 - x) which is 2 - x. Threfore, the total area A = π[(2 - x)² - x²] = π(4x + 4) = 4π(x + 1)
The equation is y = 2x - x², so
- y = x² - 2x
1 - y = x² - 2x + 1
1 - y = (x - 1)²
.: x - 1 = √(1 - y)
Sub that into A such that A = 4π√(1 - y)
To find volume, the small change in volume δV = 4π√(1 - y)δy
V = 4π∫√(1 - y) dy with limits 0 to 1 (the limiting y values of the volume)
Evaluate the integral and you should get 8π/3
For the last one, check if you got the equations in the question correct, because those two curves don't even intersect.
Last edited:
conman
Member
Thanks a lot, trebla. I think there is something wrong with question 4 'cause when I draw the graph, I could not identify with section to rotate. There is no intersection between these 2 graphs.