# Double Integration without Anti-Derivatives

1. Feb 8, 2014

1. The problem statement, all variables and given/known data
Make a good sketch of the plane region D defined by the following simultaneous inequalities:
D: y >/= -2x, 2y >/= x, 2y </= 4-x.
Use deep conceptual understanding the insight (and no antiderviative calculations!) to reduce the iterated integral below to a simple algebraic expression depending on the parameters a, b, and c:
I = ∫∫(a+bx+cy)dA.
Enter the value of I corresponding to a = -0.8, b = 0.7, and c = -1.2.

a) -7.04
b) -6.00
c) -4.67
d) 8.07
e) 2.74

2. Relevant equations
None, other than the integral equation provided in the question.

3. The attempt at a solution
As asked in the question, I made a quick sketch, shown below:

Then I used the dot product to figure out that the region in consideration is a right triangle. Hence, it was easy to calculate the area of this region in the 2D plane. I calculated it to be:
A = 3.097097656
Now, I don't know how to extend this idea to calculate the volume since f(x,y) is not constant.

Any help would be greatly appreciated. Thanks!

2. Feb 8, 2014

### haruspex

Since two of the lines are at right angles, it would be natural to do a change of co-ordinates. This will effectively turn b and c into two different coefficients, i.e. something like ∫∫(a+b'u+c'v)dA. Can you then use a geometric interpretation of ∫∫b'u.dA etc?

3. Feb 8, 2014

Unfortunately, I don't quite understand the idea that you are suggesting. I am unsure on how switching two co-ordinates will achieve anything in regards to this problem.

4. Feb 8, 2014

### SammyS

Staff Emeritus
What does the equation, z = a+bx+cy represent ?

5. Feb 8, 2014

Hm... it represents a plane. Are you suggesting that I find the volume of the two pieces (triangular base + another triangular piece on top with the slanted plane)?

6. Feb 8, 2014

### haruspex

No, SammyS and I are both suggesting you think of it as a volume integral.

7. Feb 8, 2014

### LCKurtz

I don't think your area A is correct. I suggest re-calculating it with exact values instead of decimals. Also, it isn't difficult to find the centroid of a right triangle. Your problem can be thought of as the volume of a triangular prism with slanted top.

Are you sure the values for a,b,c and the answer choices are all typed correctly?

 Never mind about the typing, all is correct except your area is only sort of close.

Last edited: Feb 8, 2014
8. Feb 8, 2014

Yes, the values are correct (as shown in my question). Althought I must point out that two previous questions have had incorrect sets of answers in this particular problem set. My professor acknowlegded them as well.

9. Feb 8, 2014

### LCKurtz

Yes, I had a mistake in calculations I found. Still, you should use exact values, certainly for the area of the triangle. And the answers do have a correct choice.

10. Feb 8, 2014

Well, I am still a bit confused as to where to go from here. After reading the responses and doing some research online, I stumbled upon something called the "Jacobian." Am I drifting off on a tangent to infinity?

11. Feb 8, 2014

### Ray Vickson

For each $(x,y) \in D$, the integrand $f = a + bx + cy$ is $f = a + \vec{u} \cdot \vec{r}$, where $\vec{u} = (c,d)$, $\vec{r} = (x,y)$ and $\cdot$ is the inner product. Since integration is a 'linear' operation, we can pull the $\vec{u} \cdot (\cdots)$ outside the integral, to get
$$\text{integral } = a \int_D\, dA + \vec{u} \cdot \int_D \vec{r} \, dA.$$
Do you recognize both of these integrals, perhaps from a physics course or somewhere similar?

12. Feb 8, 2014

No, I don't recognize this from any of my courses. I was able to follow through the logic that you presented. However, shouldn't it be $\vec{u} = (b,c)$? Also, when you break the integral into two in the end, don't I still have to use anti-derivatives for the second piece in x and y (as contained in the vector r)?

13. Feb 8, 2014

### haruspex

Not if you recognize the physical interpretation of the integral and so can apply a well-known formula.
SammyS and I were pushing you towards thinking of it as a volume integral. LCKurtz also refers to this, specifically as a triangular prism. My rotation of co-ordinates suggestion was to turn it from a triangular prism into something a bit simpler. But the best suggestion is LCKurtz's mention of centroids. (Ray V might also have that in mind.)
How do you calculate the location of the centroid of a lamina?

14. Feb 8, 2014

### Ray Vickson

Yes, it should be $\vec{u} = (b,c)$. The first term should be easy; the second term will be familiar to you if you have taken a Mechanics course; otherwise, perhaps not.

15. Feb 8, 2014

Well, for a right triangle, it's simple as the formulas are already dervided online. Or, we could simply integrate using (assuming density is constant):

x_centroid = 1/A * ∫∫x*dA
y_centroid = 1/A * ∫∫y*dA

Suppose I do find the centroid co-ordinates, how does that result into determining a volume? I feel like I am missing something big here but I just can't see it.

16. Feb 8, 2014

### LCKurtz

So you could rewrite those as$$\bar x A =\iint x~dA,~~\bar y A = \iint y~dA$$Can you use that to rewrite your problem$$I = \iint a + bx + cy~dA~\text{?}$$

17. Feb 8, 2014

### SammyS

Staff Emeritus
In the original statement of the problem, it's clear that you are not to evaluate the integral by using anti-derivatives.

I agree with LC Kurtz that you can calculate the volume of a triangular prism -- or maybe two prisms. (I haven't done the calculations.)

18. Feb 8, 2014

### LCKurtz

You don't need to do that.

19. Feb 8, 2014

### SammyS

Staff Emeritus
Oh, right!

Using the centroid takes care of all that complication !

20. Feb 8, 2014

### haruspex

No, I meant that there are two physical interpretations available for your integral: as the volume of a prism (or tetrahedron if you do the co-ordinate rotation); or as the centroid of a lamina. The second is definitely easier.