# Issue with Double Integral

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1. Oct 27, 2016

### dlacombe13

1. The problem statement, all variables and given/known data
Find the volume of the given solid:
Under the surface z = xy and above the triangle with verticies (1,1), (4,1) and (1,.2)

2. Relevant equations
Double Integral

3. The attempt at a solution
I drew the triangle, and found the the equations of the lines to be:
x = 1;
y = 1;
y = -3x + 7

My issue is, I get different answers when solving using Type 1 and Type 2. Type two gives the right answer, however Type 1 does not. I have also put my Type 1 into a solver and it also gives the wrong answer. So the error must be in the setup and not my integration, but I can't see it.

Type 2:
$\int_1^2 \int_1^{-3x+7} xy = \frac{31}{8}$

Type 1:
$\int_1^4 \int_1^{-3x+7} xy = \frac{207}{8}$

2. Oct 27, 2016

### olivermsun

First of all, be explicit with your integral notation and your dx, dy. (I don't know if there is are standard meanings for "Type 1" and "Type 2" integrals.)
Second, do you mean the third vertex of the triangle is (1, 2)? If so, then check your equation for the line by plugging in some points...

3. Oct 27, 2016

### dlacombe13

Oh shoot, wow. I realized that I used some illegal algebra which resulted in the wrong equation for the line! The equation of the line is actually supposed to be $y = -\frac{1}{3}x + \frac{7}{3}$ . Now I am getting the right answer. Thanks for questioning my equation.

4. Oct 28, 2016

### Staff: Mentor

I agree with both points olivermsum made. First, "Type 1" and "Type 2" aren't standard attributes, so these should have been explained.
Second, your integrals are missing the dx and dy parts.
From this -- $\int_1^2 \int_1^{-3x+7} xy = \frac{31}{8}$ -- we can infer that you really mean this:
$\int_1^2 \int_1^{-3x+7} xy dy dx= \frac{31}{8}$, but only because the lower limit of integration for the inner integral suggests that x is the variable for that integral.
To be really explicit, you can do this:
$\int_{x = 1}^2 \int_{y = 1}^{-3x+7} xy ~dy ~dx = \frac{31}{8}$