- #1
iScience
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if we want to find the volume of a function revolved about the x-axis all we do is find the differential element..
dV=(Area)dx =[itex]\pi[/itex]y2 , where y=y(x)
so then..
[itex]\int[/itex][itex]\pi[/itex]y2dx
the differential element looks like this...
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB
so i add a bunch of these up and i get the total volume. okay... fine..
but then we get to surface area.. and the differential element looks like this...
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#1
please don't get into the equations trying to explain to me where the frustum equation came from, i know where it came from, and i understand the rationale for using the frustum as the differential element to account for the discrepancies due to the function's slope. the trouble I am having is.. if we must choose our differential element for the surface area to be the frustum, then why is it okay to choose the cylinder as the differential element for finding the volume?
due to the fact that a differential element of a frustum takes into consideration the slope of the function, i can see why this would be more accurate/preferrable over just the side surface area of the cylinder,
ie...
[itex]\int[/itex]2[itex]\pi[/itex]ydx
umm.. here just for the hell of it i drew a picture
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#2
...
Now consider the cylindrical differential element for the volume of the rotated function
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3
(probably going to have to click to enlarge the image)
basically my question is.. if it is NOT okay to use the cylindrical differential element to find the surface area ie...
[itex]\int[/itex]2[itex]\pi[/itex]ydx
then why is it okay to use the cylindrical differential element to find the volume?
look at the image again (and enlarge it)
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3
there is leftover crap from the diff. element sagging outside all over the actual volume of the rotated function.
please help, I've had trouble with this ever since.. well.. calc II which was two years ago
dV=(Area)dx =[itex]\pi[/itex]y2 , where y=y(x)
so then..
[itex]\int[/itex][itex]\pi[/itex]y2dx
the differential element looks like this...
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB
so i add a bunch of these up and i get the total volume. okay... fine..
but then we get to surface area.. and the differential element looks like this...
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#1
please don't get into the equations trying to explain to me where the frustum equation came from, i know where it came from, and i understand the rationale for using the frustum as the differential element to account for the discrepancies due to the function's slope. the trouble I am having is.. if we must choose our differential element for the surface area to be the frustum, then why is it okay to choose the cylinder as the differential element for finding the volume?
due to the fact that a differential element of a frustum takes into consideration the slope of the function, i can see why this would be more accurate/preferrable over just the side surface area of the cylinder,
ie...
[itex]\int[/itex]2[itex]\pi[/itex]ydx
umm.. here just for the hell of it i drew a picture
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#2
...
Now consider the cylindrical differential element for the volume of the rotated function
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3
(probably going to have to click to enlarge the image)
basically my question is.. if it is NOT okay to use the cylindrical differential element to find the surface area ie...
[itex]\int[/itex]2[itex]\pi[/itex]ydx
then why is it okay to use the cylindrical differential element to find the volume?
look at the image again (and enlarge it)
http://imgur.com/HwrihKA,hjkKFzV,cpUPsvZ,YKIyTIB#3
there is leftover crap from the diff. element sagging outside all over the actual volume of the rotated function.
please help, I've had trouble with this ever since.. well.. calc II which was two years ago
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