volume of solid of revolution Watch

bigmansouf
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Question: the area enclosed by  y=x^2-6x+18  and y =10 is rotated about the x-axis; find the volume generated. take an element of area parallel to the x-axis of length (x_{2}-x_{1}) express the typical elements of volume in terms of y by using the fact that  x_{1} and  x_{2} are the roots of  x^2 -6x + (18-y)= 0

my attempt:
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 y=x^2-6x+18
when x is the subject
 x = 3+(y-9)^{0.5}


the interval is y =9 to y=10
the minimum point is (3,9)

 v= \int_{9}^{10} \pi (3+(y-9)^{0.5})^{2} \mathrm{d}y
 v = 13.5000002 \pi

please help me with this i am finding it difficult to understand how to tackle the book gives the answer as  8\pi
Last edited by bigmansouf; 6 days ago
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begbie68
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not quite sure of your method, or the method described in the question.

By a more methodical route, we can find the required volume by

vol. cylinder (radius 10, height 2) subtract volume of revolution of y=x^2-6x+18 about x-axis

I get : 25.6pi

200pi - pi*integral[x^4 - 12x^3 +72x^2 - 108x +324] , from x=2 to x=4

geogebra vol of integration tool
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old_engineer
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(Original post by bigmansouf)
Question: the area enclosed by  y=x^2-6x+18  and y =10 is rotated about the x-axis; find the volume generated. take an element of area parallel to the x-axis of length (x_{2}-x_{1}) express the typical elements of volume in terms of y by using the fact that  x_{1} and  x_{2} are the roots of  x^2 -6x + (18-y)= 0

my attempt:
the graph is Name:  a1.jpg
Views: 3
Size:  127.7 KB

 y=x^2-6x+18
when x is the subject
 x = 3+(y-9)^{0.5}


the interval is y =9 to y=10
the minimum point is (3,9)

 v= \int_{9}^{10} \pi (3+(y-9)^{0.5})^{2} \mathrm{d}y
 v = 13.5000002 \pi

please help me with this i am finding it difficult to understand how to tackle the book gives the answer as  \ pi
It looks to me that what you have tried to calculate is the volume formed by rotating the area to the left of the curve about the y axis, between y = 9 and y = 10.
begbie68 has calculated the volume mentioned in the question by starting with the volume of a disc of radius 10 and thickness 2, then subtracting the volume formed by rotating the area under the curve around the x axis. I agree with the answer 25.6(pi).
However, the question is asking you to consider a different method where the required volume is made up of a series of concentric cylindrical shells (with the x axis as the cylinder axis). Moving out from the x axis, the first cylindrical shell has radius 9 and length (x direction) 0, and the last cylindrical shell has radius 10 and length 2. You have to find a way of summing the volumes of these cylindrical shells by integration.
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