# Multiple integrals

1. May 24, 2012

### robertjford80

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I understand the steps, although it took quite a while, but what I don't understand is that a triangle with base 2 and height 2, it's area is 2. With two triangles of that size the area should be 4. The books says the area is 8.

2. May 24, 2012

### clamtrox

You see, the book is not calculating the area of the triangle, but the area times two (there's a two in the integrand)

3. May 24, 2012

### robertjford80

one point of the base is -2,-2, the other point is 0,-2 - looks like base 2 to me

4. May 24, 2012

### sharks

Consider the 2 right-angled triangles as a whole isosceles triangle. Then, the base of that isosceles triangle is 4 and the height is 2. The formula for calculating the area is still the same, meaning 1/2 x base x height = 1/2 x 4 x 2 = 4. Multiply that 4 by 2 (from the integrand) and you get the answer = 8.

Now, if you use the double integral given, you should get the same answer.

5. May 24, 2012

### robertjford80

That to me is the answer.

Why are you doing that?

6. May 24, 2012

### sharks

OK, now that you agree that the area of the whole triangle is 4, we go back to look at your problem, which is:
$$\iint 2\,.dpdv$$
Ignore the limits (just to simplify your understanding). Notice the "2". That's what we call the integrand. Note that the integrand is a constant in this case. The integrand could have been anything. As a general rule, all constants in integrands can be "pulled out" of the integration. So, we get the following which is equivalent:
$$2\iint \,.dpdv$$
The area of the entire triangle in the figure is given by:
$$\iint \,.dpdv$$
And you have seen that the area of the entire triangle is 4. What does this mean? It just means that:
$$\iint \,.dpdv=4$$
If you just put it all back together:
$$\iint 2\,.dpdv=2\iint \,.dpdv=2\times 4 =8$$

7. May 24, 2012

ok, thanks.