# How do you determine the area to the left of g(y)=3-y^2 and to the right of x=-1?

Mar 27, 2015

Determine the area to the left of $g \left(y\right) = 3 - {y}^{2}$ and to the right of $x = - 1$

Turn your thinking ${90}^{\circ}$.

This is the dual problem of the problem: Find the area under $f \left(x\right) = 3 - {x}^{2}$ and above $y = - 1$.

in terms of $y$ "left" = in terms of $x$ "under"

in terms of $y$ "right" = in terms of $x$ "above"

You don't need to change the problem if you can change your thinking. Usually (initially) we found the area between two functions ox $x$ by integrating $\int$ (the top function - the bottom function) $\mathrm{dx}$

This time we'll do:

$\int$ (right function - left function) $\mathrm{dy}$

Limits of integration are the values of $y$ at the points of intersection. Namely: solve the system: $x = 3 - {y}^{2}$ and $x = - 1$.

$3 - {y}^{2} - - 1$ at $y = \pm 2$

So now we know to evaluate:

int_-2^2((3-y^2-(-1)) dy int_-2^2(4-y^2)dy which I think you can finish.