How to change order of integration in a double integral?

[luminousxx0]

Could you help? is it a wrong question

 

[Deleted member 4993]

Could you help? is it a wrong question
View attachment 18417

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this assignment.

If I were to do this problem:

The limits need to be changed. Thus I would plot y = -3x and y = 4-x² to decide on new limits.

 

[luminousxx0]

is that correct until last step? i didn't continue this step.

 

[HallsofIvy]

The first thing I would do is graph the equations. The lower boundary, y= -3x, is a straight line through the origin with slope -3, the upper boundary, y= 4- x^2, is a parabola with vertex at (0, 4) opening downward. The second thing I would do is determine the coordinates of the points where those boundaries intersect. y= -3x= 4- x^2 gives x^2- 3x- 4= 0. That factors easily as (x- 4)(x+ 1)= 0 so they intersect at (-1, 3) and (4, -12).

Further I see that if I were to draw horizontal lines across that region, corresponding to different values of y, the left and right end points, the limits on the x integrals, change. For y less than or equal to 3, the left end point is on the line y= -3x. For y greater than or equal to 3, the left endpoint is on the left side of the parabola y= 4- x^2. That tells me that I have to break the integral into two parts, one for y from -12 to 3 and the other for y from 3 to 4.

The line y= -3x "inverts" to x= -y/3. The parabola y= 4- x^2 doesn't have a true "inverse". x^2= 4- y so x= (4- y)^(1/2), the right branch, or x= -(4- y)^(1/2), the left branch.

Reversing the order of integration we need to break this into two separate double integrals:
\(\displaystyle \int_{y= -12}^3 \int_{x= -y/3}^{\sqrt{4- y}} f(x,y) dxdy+ \int_{y= 3}^4\int_{x= -\sqrt{4- y}}^{\sqrt{4- y}} f(x,y) dxdy\).

 

[luminousxx0]

Thanks for taking the time to help me.
Did I draw the graph correct? I tried to calculate area but i couldn't do it.
Could you help me?

 

[Cubist]

Did I draw the graph correct? I tried to calculate area but i couldn't do it.
Could you help me?

Your graph is correct, well done!

HallsofIvy has provided the answer, which I think is correct. I just wanted to post an image to show how the original, and reversed, integrals proceed within the area. Please post back if you're confused by any of this.

 

[Steven G]

Just look at the (horizontal) strips which Cubist drew for you. Is it true that over the entire region that one curve is always above the other curve (yes, a line is a curve). If not then you will need to sum up two or more integrals.

 

[luminousxx0]

A=125/6 finished