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.

amistre64 · Answer

|dw:1329597568790:dw|

amistre64 · Answer

the revolution is simply an area of a circle; pi [f(x)]^2

anonymous · Answer

Can you please explain in depth

amistre64 · Answer

$$pi\int_{0}^{2}(e^{2x}-e^{-2x})dx$$
seems to be what it amounts to then

amistre64 · Answer

in depth; which part?

amistre64 · Answer

|dw:1329597700997:dw|

anonymous · Answer

I have to use big R and little r both squared because the shape is hollow right? so I take both the solid and hollow part as my big R squared and then just the hollow part as my little r squared?

amistre64 · Answer

the radius of the circle whose area we want is determined by the function f(x)  correct?

amistre64 · Answer

yes, Big radius little radius

anonymous · Answer

Thats what you put in the integral right?

amistre64 · Answer

yes$$R_1 ^2=(e^x)^2 = e^2x$$$$R_2 ^2=(e^{-x})^2=e^{-2x}$$

anonymous · Answer

okay thank you !

amistre64 · Answer

if you get a negative result; dont worry, it just means you got your top and bottom function swapped; drop the negative and your answer will be good

anonymous · Answer

thanks :D

amistre64 · Answer

youre welcome :)