Double integral to find area

[fishturtle1]

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..

 

[Mark44]

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.

You said you drew a sketch of the region, but your integral doesn't take into account that half of the region is below the x-axis. Both graphs are parabolas: one opens to the left and one opens to the right.
You have two mistakes in the first integral below:

1. Part of the region is below the x-axis.
2. The y-coordinate at the intersection point isn't 2.

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.

h(x) = 1 when you're find the area of the region of integration.

 

[Charles Link]

For the limits on your $y$ integral, you need to first determine where these two curves intersect. $y=0$ and $y=2$ is incorrect.

 

[fishturtle1]

You said you drew a sketch of the region, but your integral doesn't take into account that half of the region is below the x-axis. Both graphs are parabolas: one opens to the left and one opens to the right.
You have two mistakes in the first integral below:

1. Part of the region is below the x-axis.
2. The y-coordinate at the intersection point isn't 2.

For the limits on your yy y integral, you need to first determine where these two curves intersect. y=0y=0 y=0 and y=2y=2 y=2 is incorrect.

I drew $y = x^2$ and $y = 2 - x^2$ which I see is wrong.
This time I drew $x = y^2$ and $x = 2 - y^2$.

I see the two intersections of the curves are at (1,1) and (1,-1). So my limits of integration for the y integral are -1 to 1. So,

Area = $\int_{-1}^{1} \int_{y^2}^{2-y^2} 1 dx dy$..I put the 1 in there just to remember its there,
= $\int_{-1}^{1} 2-2y^2 dy$
= $(2y - \frac{2y^3}{3}) |_{-1}^{1}$
=$(2 - \frac 23) - (-2 + \frac 23)$
=$(\frac {12}{3} - \frac 43)$
=$\frac 83$

Thank you for the help.

 

[Mark44]

Regarding the 1 in the integrand, one application of double integrals is finding the volume of a three-d region in space. For your problem, you can think of it as a solid whose base is that region between the two parabolas, and whose height is 1. The area of the region and the volume of the solid are numerically equal, but of course, one is an area and the other is a volume, so the units would be different.

 

[fishturtle1]

Regarding the 1 in the integrand, one application of double integrals is finding the volume of a three-d region in space. For your problem, you can think of it as a solid whose base is that region between the two parabolas, and whose height is 1. The area of the region and the volume of the solid are numerically equal, but of course, one is an area and the other is a volume, so the units would be different.

Thanks that makes sense