# find the area of surface generated by revolving the are about y axis

#### HallsofIvy

MHF Helper
?? You don't want to find t such that $$\displaystyle x= e^tcos(t)= 0$$ and $$\displaystyle x= e^t cos(t)= 1$$! The problem specifically says that it is y that goes from 0 to 1. $$\displaystyle y= e^t sin(t)= 0$$ for t= 0 and $$\displaystyle y= e^t sin(t)= 1$$ when x is about -3.6.

#### xl5899

?? You don't want to find t such that $$\displaystyle x= e^tcos(t)= 0$$ and $$\displaystyle x= e^t cos(t)= 1$$! The problem specifically says that it is y that goes from 0 to 1. $$\displaystyle y= e^t sin(t)= 0$$ for t= 0 and $$\displaystyle y= e^t sin(t)= 1$$ when x is about -3.6.
other than that , is my other working correct ?

#### xl5899

?? You don't want to find t such that $$\displaystyle x= e^tcos(t)= 0$$ and $$\displaystyle x= e^t cos(t)= 1$$! The problem specifically says that it is y that goes from 0 to 1. $$\displaystyle y= e^t sin(t)= 0$$ for t= 0 and $$\displaystyle y= e^t sin(t)= 1$$ when x is about -3.6.
other than that , is my other working correct ?

MHF Helper

#### xl5899

can you point out the mistake ? i couldnt discover it after working on it almost half day

#### Prove It

MHF Helper
Look at your \displaystyle \begin{align*} \left( \frac{\mathrm{d}x}{\mathrm{d}t} \right) ^2 \end{align*} term...

• 1 person

#### xl5899

Look at your \displaystyle \begin{align*} \left( \frac{\mathrm{d}x}{\mathrm{d}t} \right) ^2 \end{align*} term...
whats wrong with it ?

#### xl5899

Look at your \displaystyle \begin{align*} \left( \frac{\mathrm{d}x}{\mathrm{d}t} \right) ^2 \end{align*} term...
can you point out the mistake directly ?