Solving a Double Integral: Int[1 to 2] Int[0 to x] of 1/(sqrt((x^2)+(y^2)))dydx

  • Thread starter FecalFrown
  • Start date
The final result is \left.\frac{1}{x}\ln\left(\frac{x+\sqrt{x^2+y^2}}{y}\right)\right|_0^x=\frac{1}{x}\ln\left(\frac{2x}{x}\right)=\ln(2).In summary, the problem involves solving a double integral with a trigonometric substitution. The final result is ln(2).
  • #1
FecalFrown
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Homework Statement


Howdy fellas, long time lurker, first time poster. If someone could get me started on the right track for this problem, it would be appreciated.

Also, I'm not sure how to write equations yet, but I'll do my best to explain the problem clearly

Homework Equations



Int[1 to 2] Int[0 to x] of 1/(sqrt((x^2)+(y^2)))dydx

The Attempt at a Solution



As I said, not sure how to get started. My gut was telling me to raise the denominator to the -1/2 power and use the substitution rule? But I'm not sure.

Thanks
 
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  • #2
Hi FecalFrown! :smile:

Let's start with the inner integral, this is

[tex]\int_0^x{\frac{1}{\sqrt{x^2+y^2}}dy}[/tex]

where x is simply a constant. The trick is to factor x out of the square root, and then to do a trigonometric substitution [itex]\frac{y}{x}=\tan(\theta)[/itex]
 

FAQ: Solving a Double Integral: Int[1 to 2] Int[0 to x] of 1/(sqrt((x^2)+(y^2)))dydx

What is a double integral?

A double integral is a type of mathematical operation that involves evaluating a function over a two-dimensional region. It is often used in calculus and physics to calculate areas, volumes, and other quantities.

Why is this specific double integral important?

This particular double integral, Int[1 to 2] Int[0 to x] of 1/(sqrt((x^2)+(y^2)))dydx, is important because it represents the area under a specific curve in a two-dimensional plane. This curve is known as the inverse tangent function and is used in many applications in mathematics and physics.

How do you solve a double integral?

To solve a double integral, you must first determine the limits of integration, which define the boundaries of the region over which the function will be evaluated. Then, you must set up the integral using the appropriate notation and evaluate it using techniques such as substitution or integration by parts.

What is the relationship between a double integral and a single integral?

A double integral is essentially the combination of two single integrals, one nested within the other. The outer integral is often used to determine the limits of the inner integral, which is then evaluated to find the final result.

What are some real-world applications of double integrals?

Double integrals have many real-world applications, such as calculating the center of mass of a two-dimensional object, finding the area between two curves, and determining the volume of a three-dimensional object. They are also used in physics to calculate electric and magnetic fields, as well as in probability and statistics to find the joint probability of multiple events.

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