Question

Can you help me with this double integral?

`intint_D (y^2-x)dA` where D is limited by y=x-6 , `y^2=x`

Answer 1

Please see below.

Explanation:

Sketch the region `D`, shown below.

Now describe the region so that we can integrate either w.r.t `x` first and then `y` or integrate first w.r.t. `y`, the `x`.

We can describe the region in either of two ways.

Method 1
One description notes that we can integrate with respect to `x` second, using constant limits of i=ntegration `0` and `9`.
But not really.

As `x` varies from `0` to `4`, the values of `y` go from the lower half parabola (`y = -sqrtx`) to the upper half parabola (`y = sqrtx`).
The bounds on `y` are different as `x` varies from `4` to `9`

This lead to the two double integrals

`int_0^4 int_(-sqrtx)^(sqrtx)(y^2-x) \ dy \ dx + int_4^9 int_(x-6)^(sqrtx) (y^2-x) \ dy \ dx`

Method 2
I think it is simpler to note that `y` varies from `-2` to `3`,
and on that entire interval, `x` varies from the parabola (`x=y^2`) to the line (`x=y-6`)

This allows us to use the one double integral:

`int_(-2)^3 int_(y^2)^(y+6) (y^2-x) \ dx \ dy`.

Answer 2

I believe the following to be a correct way to do this integration but I will request a double check so that one of my peers will review it.

Explanation:

Express the limits as x in terms of y:

`x = y^2, x = y+6`

The following is a graph of the area:

www.desmos.com

In terms of x, `y^2` is the lower limit, therefore, the integration with respect to x is from `y^2` to `y+6`:

`int int_(y^2)^(y+6) (y^2-x)dxdy`

There are two ways to find the limits of y, solve the quadratic that results from setting `y^2` equal to `y + 6`, or observe the y coordinate where the line insects the parabola; I shall do the former:

`y^2 = y+6`

`y^2 -y - 6 = 0`

`(y+2)(y-3) = 0`

`y=-2 and y=3`

Please observe that the graph confirms this result.

The integration with respect to y is from `-2` to `3`:

`int_-2^3 int_(y^2)^(y+6) (y^2-x)dxdy`

Integrate with respect to x:

`int_-2^3 {:y^2x-1/2x^2|_(y^2)^(y+6)dy`

Evaluate at the limits:

`int_-2^3 y^2(y+6)-1/2(y+6)^2-y^2(y^2)+1/2(y^2)^2dy`

Simplify:

`int_-2^3 -y^4/2 + y^3 + (11 y^2)/2 - 6 y - 18dy`

Integrate:

`-y^5/10 + y^4/4 + (11 y^3)/6 - 3 y^2 - 18 y |_-2^3`

Evaluate at the limits:

`-(3)^5/10 + (3)^4/4 + (11 (3)^3)/6 - 3 (3)^2 - 18 (3)- (-(-2)^5/10 + (-2)^4/4 + (11 (-2)^3)/6 - 3 (-2)^2 - 18 (-2)) = -625/12`