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Wednesday, January 27, 2010

Q.108. Volume of a solid of revolution.

Question 108.
Find the volume of the solid obtained by rotating the region enclosed by the curves
y = 16 - x, y = 3x + 12, x = 0 about y axis using disk method.

Answer 108.
For the equations, y = 16 - x, y = 3x + 12, x = 0, 
the points of intersection of the lines are A(0, 12), B(0, 16) and C(1,15).

It is a small triangle ABC which is to be rotated about y-axis.
Shifting origin to (0, 12), new coordinates of ABC will be (0, 0), (0, 4) and (1, 3)
and the new equations of lines will be y = 3x and y = 4 - x

=> Required area
= π ∫ (y=0 to 3) y^2/9 dy + π ∫ (4 - y)^2 dy
= (π/9) (y^3/3) [y=0 to 3] + π (16y - 4y^2 + y^3/3) [y=3 to 4]
= π + π (64/3 - 21)
= 4π/3.
 [This can easily be verified to be true by treating the required volume as the sum of volumes of two cones, one of radius 1 and height 3 and the other of radius 1 and height 1.]

LINK to YA!

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