Solving Multiple Integrals - Understanding Triangle Area Calculation

In summary, the book says the area of a triangle is 8, but the area of the triangle when calculated using the integrand is 2.
  • #1
robertjford80
388
0

Homework Statement



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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.
 
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  • #2
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
one point of the base is -2,-2, the other point is 0,-2 - looks like base 2 to me
 
  • #4
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
sharks said:
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?
 
  • #6
robertjford80 said:
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:
[tex]\iint 2\,.dpdv[/tex]
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:
[tex]2\iint \,.dpdv[/tex]
The area of the entire triangle in the figure is given by:
[tex]\iint \,.dpdv[/tex]
And you have seen that the area of the entire triangle is 4. What does this mean? It just means that:
[tex]\iint \,.dpdv=4[/tex]
If you just put it all back together:
[tex]\iint 2\,.dpdv=2\iint \,.dpdv=2\times 4 =8[/tex]
 
  • #7
ok, thanks.
 

FAQ: Solving Multiple Integrals - Understanding Triangle Area Calculation

What is a multiple integral?

A multiple integral is an extension of a single integral to integrate functions of more than one variable over a specified region in space. It can be thought of as the limit of a sum of areas of rectangles in a plane as the number of rectangles approaches infinity.

How do you solve a multiple integral?

To solve a multiple integral, you must first determine the limits of integration for each variable. Then, you can use the appropriate integration techniques, such as substitution or integration by parts, to evaluate the integral. Finally, you can combine the results to find the overall solution.

What is the purpose of understanding triangle area calculation in multiple integrals?

Understanding triangle area calculation in multiple integrals is important because it allows you to accurately calculate the area of a three-dimensional object or region. It also serves as a fundamental building block for more complex integrals and can be applied to other shapes and geometries.

How do you calculate the area of a triangle using multiple integrals?

The area of a triangle can be calculated using the formula A = 1/2 * base * height. In the case of multiple integrals, the base and height are represented as functions of two variables, and the integral is taken over the region of the triangle. The result is the area of the triangle.

What are some common applications of multiple integrals in science?

Multiple integrals have various applications in science, including calculating volumes of three-dimensional shapes, computing the center of mass of an object, and finding the average value of a function over a region. They are also used in fields such as physics, engineering, and economics to model and solve real-world problems.

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