Solving Multiple Integrals - Understanding Triangle Area Calculation

[robertjford80]

Homework Statement

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.

 

[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)

 

[robertjford80]

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

 

[DryRun]

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.

 

[robertjford80]

The formula for calculating the area is still the same, meaning 1/2 x base x height = 1/2 x 4 x 2 = 4.

That to me is the answer.

Multiply that 4 by 2 (from the integrand) and you get the answer = 8.

Why are you doing that?

 

[DryRun]

Why are you doing that?

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

 

[robertjford80]

ok, thanks.