volume problem 2
bounds y=ln x to 1
just switch the x and y in the Y=lnx and integrate from 0 to 1 since there is no x axis limits but there is one on the y axis
1-y = x?
\[\pi \int\limits_{0}^{1} (e^y)^2 dy\]
you want your equation to be in the form x= so solve for x in the equation
where did the e come from tho
ohhhh cuz of ln right
the equation is y=lnx right? so solve for x now so exponentiate both sides e^y=e^lnx so the e and the ln cancel out on the right and you are left with e^y=x
ok ok yeah yeah blonde moment hahha
yea so now that you are in x=e^y form you integrate up and down not left and right so it would be the integral i posted above
well i just put that whole equation in my calc n i got a different answer then the back
\(2\pi\int\limits_{1}^{e}x\cdot(1-\ln(x))\;dx\) Keep your mind open to other possibilities.
wait dakota the answer is supposed to be 10. something
did u multiply by pi? lol
Note: This supports and verifies dakotajason27's response if you remember to add the additional \(\pi\) that I forgot. It's about 10.036.
oh shoot yea thanks :D
haha no problem :b
countless times have i forgot the pi :/
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