- #1

fishturtle1

- 394

- 82

## Homework Statement

Use double integrals to find the areas of the region bounded by ##x = 2 - y^2## and ##x = y^2##

## Homework Equations

Volume = ##\int_a^b \int_{f(x)}^{g(x)} h(x) dx dy##.. and this is equivalent if I switched the integrals and redid the limits of integration

## The Attempt at a Solution

[/B]

I am pretty sure I did this wrong, so I'm asking for any help to do it right.

Area = ##\int_0^2 \int_{y^2}^{2-y^2} dx dy##

=##\int_0^2 x |_{y^2}^{2-y^2}dy##

=##\int_0^2 2-2y^2 dy##

=##(2y - \frac {2y^3}{3}) |_0^2##

=##4 - \frac {16}{3} = 4 - 5\frac 13 = -\frac 43##

I think this answer is wrong because it is negative, and from the graph I drew, the area should be greater than 2. I'm not sure what to do about the h(x) in the relevant equations. If I set it to 0, then Area = 0, but if I set it to any constant then I am solving for a volume.

edit: looking at this more, I'm pretty sure ##\int_{y^2}^{2-y^2} dx## is not right..

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