Find the exact area of the region enclosed by the curve given by $$x=9-t^2$$ $$y=e^t$$ where $-3 \leq t \leq 3$ and the $y$-axis.
I tried to take the integral of the $x$ function minus the $y$ function times $dx$, but then I got stuck at that point. Any help would be greatly appreciated.
On
Note that $x=0$ when $t=\pm3$ and that at $t=\pm3$, $y=e^{-3}$ or $y=e^3$. At $t=0$ $x=9$ and $y=1$. Also, solving for $x$ in terms of $y$ gives $x=9-\ln^2y$. Thus you must evaluate $\int_{e^{-3}}^{e^3}9-\ln^2y\,dy$ which requires integration by parts but is straightforward.
Millikan is right, though, it is easier in terms of $t$.
$\int x\,dy$ with $dy=e^t\,dt$ gives $\int_{-3}^3(9-t^2)e^t\,dt$ which still requires integration by parts but is much easier.
Hint: To find an area, you integrate $\int y\; dx$ or (in this case) $\int x \;dy$. Intuitively, you are adding up the area of rectangles. Here $x$ is the height above the $y$ axis and $dy$ is a small extent in $y$. You need to express $x$ as a function of $y$ or else to use the chain rule $dy=\frac {dy}{dt}dt$ and integrate over $t$. The second seems easier.