Question

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

Answer 1

Determine the area to the left of `g(y)=3-y^2` and to the right of `x=-1`

Turn your thinking `90^@`.

This is the dual problem of the problem: Find the area under `f(x)=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) `dx`

This time we'll do:

`int` (right function - left function) `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=+-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.