Finding the Equation of the Hypotenuse of a Triangle in Multi-Variable Calculus

In summary, the conversation discusses finding the bounds for a double integral in a triangular region with vertices at (0,0), (5,0), and (0,3). The correct bounds would not be 0 to 5 for x and 0 to 3 for y, as those would be the bounds for a rectangular region. The equation of the hypotenuse of the triangle is x^2+y^2=sqrt(34), and the correct bounds for x would be 0 to sqrt(sqrt(34)-y^2) while the bounds for y would be 0 to 3.
  • #1
Larrytsai
228
0

Homework Statement


Evaluate the double integral I=int(int(D)( xydA) where D is the triangular region with vertices (0,0)(5,0)(0,3).


The Attempt at a Solution



I was wondering if my bounds for x would be 0 to 5 and y to be 0 to 3
 
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  • #2
Larrytsai said:

Homework Statement


Evaluate the double integral I=int(int(D)( xydA) where D is the triangular region with vertices (0,0)(5,0)(0,3).


The Attempt at a Solution



I was wondering if my bounds for x would be 0 to 5 and y to be 0 to 3
No they wouldn't. If the region were a rectangle, those would be the bounds, but your region is a triangle. Think about the equation of the hypotenuse of the triangle.
 
  • #3
k so my hypotnuse is x^2+y^2 = sqrt(34), then i isolate for x and have x=sqrt(sqrt(34)-y^2) so my bounds for x would be 0 to x=sqrt(sqrt(34)-y^2) , and y would be 0 to 3
 
  • #4
Larrytsai said:
k so my hypotnuse is x^2+y^2 = sqrt(34), then i isolate for x and have x=sqrt(sqrt(34)-y^2) so my bounds for x would be 0 to x=sqrt(sqrt(34)-y^2) , and y would be 0 to 3
That is NOT the equation of the hypotenuse of the triangle - it's the equation of a circle centered at the origin.

The hypotenuse of the triangle is a straight line segment. Do you know how to find the equation of a line?
 

1. What is multi-variable calculus?

Multi-variable calculus is a branch of mathematics that deals with functions of multiple variables, typically two or three. It is an extension of single variable calculus and involves the study of rates of change and the behavior of functions with multiple variables.

2. What are the applications of multi-variable calculus?

Multi-variable calculus has many real-world applications, including in physics, engineering, economics, and statistics. It is used to model and analyze systems with multiple variables, such as motion of objects in three-dimensional space, optimization problems, and probability distributions.

3. What are the key concepts in multi-variable calculus?

The key concepts in multi-variable calculus include partial derivatives, multiple integrals, vector calculus, and optimization. Partial derivatives are used to calculate the rate of change of a function with respect to one variable while holding the others constant. Multiple integrals are used to find the volume under a surface in three-dimensional space. Vector calculus involves the study of vector fields and line integrals, and optimization is the process of finding the maximum or minimum value of a function.

4. How is multi-variable calculus different from single variable calculus?

Multi-variable calculus differs from single variable calculus in that it deals with functions of multiple variables, whereas single variable calculus deals with functions of a single variable. This means that in multi-variable calculus, there are multiple variables that can change and affect the behavior of the function, making it more complex. Additionally, multi-variable calculus introduces new concepts such as partial derivatives, multiple integrals, and vector calculus.

5. What are the prerequisites for learning multi-variable calculus?

The prerequisites for learning multi-variable calculus include a strong understanding of single variable calculus, including concepts such as limits, derivatives, and integrals. It is also helpful to have a good grasp of algebra, trigonometry, and geometry. Some familiarity with vectors and matrices may also be beneficial.

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