Let R be the region in the xy-plane between the graphs of y = ex and y = e-x from x = 0 to x = 2.
a) Find the volume of the solid generated when R is revolved about the x-axis.

Question

Let R be the region in the xy-plane between the graphs of y = ex and y = e-x from x = 0 to x = 2.
	a) Find the volume of the solid generated when R is revolved about the x-axis.

roadjester · Answer

You have a solid of revolution. Therefore you need to take an integral. Your lower limit of integration is 0 (which makes the eventual evaluation easier) and an upper limit of integration of 2. Your going to revolve around the x-axis so you don't need to worry about shifts. Since you have it revolving around the x-axis, you have a (I can't recall the name but it isn't the washer method) easy integration.

Keep in mind the intersection point of the two graphs when you evaluate. You will need to split the integrals with respect to dx. The first set of limits is 0-->c, the second is c-->2 where c is the intersection of the two functions.

roadjester · Answer

Wow that was long.

amistre64 · Answer

there is no intersection

roadjester · Answer

did you graph the two functions?

amistre64 · Answer

this is spose to read e^(x)  and e^(-x)

amistre64 · Answer

http://openstudy.com/users/amistre64#/updates/4f4ab2fee4b00c3c5d3385c0 the posters classmates have all tried it out so far

roadjester · Answer

we'll let tserio94 decide that.

anonymous · Answer

its suppose to be e^x and e^-x

roadjester · Answer

then why did you write ex and e-x?

amistre64 · Answer

but yes, its commonly refered to as the washer method :)

anonymous · Answer

i didnt realize it, i just copied it how it was on the file the problem was on

roadjester · Answer

In that case, do you still need help?

anonymous · Answer

yeah and how would i be able to show this on a graph?

roadjester · Answer

Well, e^x is an exponential function and e^(-x) is it's reflection about the y axis. Since it's exponential, e^(-x) can also be inverted and written as 1/(e^x) but that is kind of pointless unless you prefer that notation.

roadjester · Answer

Plus, e^x or ANY exponential function, unless shifted, will pass through the point (0,1) so since your limits of integration are 0 and 2, you can really just ignore e^x.

roadjester · Answer

Does that make sense? My script crashed so I can't use the equation editor or anything.

turingtest · Answer

|dw:1331177330770:dw|this is the graph
the region is $$R=\int_{0}^{2}e^x-e^{-x}dx$$remembering that it is going around the x-axis, we want rings of outer radius e^x and inner radius e^(-x)