# Thread: finding area and volume

1. ## finding area and volume

the area enclosed between the curve y=e^x ,the x axis and the lines x=0 and x=2 is revolved about the x axis . find the volume swept out using intergration
2
i know that , PI ( ex^2 dx
)
0
the promblem is with the e^x to the power 2 if i intergrate it do i have to treat it as 2 functions and if so how do i do that cause never done that in intergration only in differentation. ty for any help

2. Originally Posted by alpharebel10
the area enclosed between the curve y=e^x ,the x axis and the lines x=0 and x=2 is revolved about the x axis . find the volume swept out using intergration
2
i know that , PI ( ex^2 dx
)
0
the promblem is with the e^x to the power 2 if i intergrate it do i have to treat it as 2 functions and if so how do i do that cause never done that in intergration only in differentation. ty for any help
$\displaystyle \left( e^x \right)^2 = e^{2x}$ NOT $\displaystyle e^{x^2}$.

3. $\displaystyle \int_{0}^{2} (e^x)dx$ this is the area

$\displaystyle \pi\int_{0}^{2} (e^x)^2 dx$ this is the volume

$\displaystyle \pi\int_{0}^{2} (e^{2x}) dx$

try to lean latex it is easy

thats what I write in latex but you should delete the 1 in the [1/math]
then you will have the same as what I write

[tex]\int_{0}^{2} (e^x)dx[1/math]

[tex]\pi\int_{0}^{2} (e^x)^2 dx[1/math]

[tex]\pi\int_{0}^{2} (e^{2x}) dx[1/math]

4. Originally Posted by Amer
$\displaystyle \int_{0}^{2} (e^x)dx$ this is the area

$\displaystyle \pi\int_{0}^{2} (e^x)^2 dx$ this is the volume

$\displaystyle \pi\int_{0}^{2} (e^{2x}) dx$

try to lean latex it is easy

thats what I write in latex but you should delete the 1 in the [1/math]
then you will have the same as what I write

[tex]\int_{0}^{2} (e^x)dx[1/math]

[tex]\pi\int_{0}^{2} (e^x)^2 dx[1/math]

[tex]\pi\int_{0}^{2} (e^{2x}) dx[1/math]
thanks for your help but has the e^2x been intergrated or do i have to do that now then just plug the numbers in

5. no you have to integrate it you see the integrate symbol right