# Setting up a double integral

• October 12th 2010, 05:39 PM
offycakes
Setting up a double integral
Hello,

I have been thinking about this double integral for a couple hours and I can't seem to recall how to properly set it up.

I am attempting to integrate something under the surface of z = 1 / (y+2) and over the area which is bounded by y=x and y^2+x=2.

I initially figured that this integral would be simpler if you integrated with respect to 'x' first. Making the lower bound for x, y; and the upper bound 2-y^2. From this I got the y lower boundary to be -sqrt 2 and upper bound sqrt 2. I have attempted multiple variations and cannot seem to arrive at the correct answer.

Can anyone assist me?
• October 12th 2010, 06:33 PM
mr fantastic
Quote:

Originally Posted by offycakes
Hello,

I have been thinking about this double integral for a couple hours and I can't seem to recall how to properly set it up.

I am attempting to integrate something under the surface of z = 1 / (y+2) and over the area which is bounded by y=x and y^2+x=2.

I initially figured that this integral would be simpler if you integrated with respect to 'x' first. Making the lower bound for x, y; and the upper bound 2-y^2. From this I got the y lower boundary to be -sqrt 2 and upper bound sqrt 2. I have attempted multiple variations and cannot seem to arrive at the correct answer.

Can anyone assist me?

The solution to $x^2 + x = 2$ is x = 1 and x = -2.
• October 12th 2010, 08:56 PM
offycakes
I did consider that, but to clarify...

Would the bounds for x be -2, 1 (lower/upper)
Bounds for y be 0 and sqrt(2-y) ?
• October 12th 2010, 11:03 PM
mr fantastic
Quote:

Originally Posted by offycakes
I did consider that, but to clarify...

Would the bounds for x be -2, 1 (lower/upper)
Bounds for y be 0 and sqrt(2-y) ?

I thought you were going to integrate with respect to x first? Have you drawn the region of integration in the xy-plane and labelled the coordinates of the intersection points of the line and parabola?
• October 13th 2010, 08:39 AM
offycakes
Yes, well my initial thought was to set x=0 to determine the y bounds. y^2+0=2, y=sqrt(2). I have also thought about the parabola being greater than y=-2, since if it was the problem would indefinite as a result of the denominator. I am still somewhat at a loss of how to progress further.
• October 13th 2010, 03:45 PM
offycakes
I finally figured the problem out, the bounds for x were not as you wrote them. The bounds for x are y and 2-y^2 (lower, upper).

The bounds for y are -2 and 1 (lower, upper).

This yields the correct result of 9/2.
• October 13th 2010, 04:10 PM
mr fantastic
Quote:

Originally Posted by offycakes
I finally figured the problem out, the bounds for x were not as you wrote them. The bounds for x are y and 2-y^2 (lower, upper).

The bounds for y are -2 and 1 (lower, upper).

This yields the correct result of 9/2.

I never said that the bounds for x were -2 and 1. I thought it would be obvious, especially at this level of question, that in my first reply the solutions I gave were for the x-coordinates of the intersection of the line and the parabola, from which the correct y-coordinates could be got. I thought my second reply made that particularly clear.
• October 13th 2010, 04:31 PM
offycakes