The idea here is that we want to take very thin "cylindrical shells" parallel to the axis we are rotating around. These shells will have a height given by functions of x that bound it and a radius given by distance from the axis of revolution (in this case y). We then just have to use the formula for the surface area of the cylinder (without filling in the "top" and "bottom," as we just have shells) given by 2Trh and integrate with respect to giving very small widths to our cylindrical shells (can you see why in this case it will be a "dy" integral?). Getting to the math, we find that the height of our cylinders is bounded by the base of the cone to where perpendicular lines from the base hit the top side of the cone. Hence, our height is given by h – x. As noted before, our radius is simply the distance of these heights from the axis of revolution which is y. But, now we seem to have a problem, we wanted a "dy" integral. x. We then get that h – "y is our height and we can get to the integral (now from y = 0 to y = r) to be What we can do use the fact that x = use this function instead of y -

Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter65: Achievement Review—section Six
Section: Chapter Questions
Problem 44AR: Solve these prism and cylinder exercises. Where necessary, round the answers to 2 decimal places...
Question
The idea here is that we want to take very thin "cylindrical shells" parallel to the axis we are
rotating around. These shells will have a height given by functions of x that bound it and a
radius given by distance from the axis of revolution (in this case y). We then just have to use the
formula for the surface area of the cylinder (without filling in the "top" and "bottom," as we just
have shells) given by 27Trh and integrate with respect to giving very small widths to our
cylindrical shells (can you see why in this case it will be a "dy" integral?).
Getting to the math, we find that the height of our cylinders is bounded by the base of the cone
to where perpendicular lines from the base hit the top side of the cone. Hence, our height is
given by h – x. As noted before, our radius is simply the distance of these heights from the
axis of revolution which is y. But, now we seem to have a problem, we wanted a "dy" integral.
5x. We then
0 to y = r) to be
So Surface Area(cylindrical shells) dy which becomes f 27(y)(h – y)dy. Evaluate
What we can do is use the fact that x =
"y and use this function instead of y
get that h – y is our height and we can get to the integral (now from y
-
this to show it is the same as the formula you got in 4.1.
Transcribed Image Text:The idea here is that we want to take very thin "cylindrical shells" parallel to the axis we are rotating around. These shells will have a height given by functions of x that bound it and a radius given by distance from the axis of revolution (in this case y). We then just have to use the formula for the surface area of the cylinder (without filling in the "top" and "bottom," as we just have shells) given by 27Trh and integrate with respect to giving very small widths to our cylindrical shells (can you see why in this case it will be a "dy" integral?). Getting to the math, we find that the height of our cylinders is bounded by the base of the cone to where perpendicular lines from the base hit the top side of the cone. Hence, our height is given by h – x. As noted before, our radius is simply the distance of these heights from the axis of revolution which is y. But, now we seem to have a problem, we wanted a "dy" integral. 5x. We then 0 to y = r) to be So Surface Area(cylindrical shells) dy which becomes f 27(y)(h – y)dy. Evaluate What we can do is use the fact that x = "y and use this function instead of y get that h – y is our height and we can get to the integral (now from y - this to show it is the same as the formula you got in 4.1.
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