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

2 Answers
Jun 11, 2018

Please see below.

Explanation:

Sketch the region D, shown below.

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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.

Jun 11, 2018

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.comwww.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