# Double Integration (Stuck at square root step) (Solution Included).

• s3a
In summary, the problem involves double integration with an additional ρ under the square root. The attempt at a solution includes using trigonometry with ρ = 2acosθ or ρ = 2asinθ, but it is unclear which to choose and whether ρ should be replaced with a different variable when integrating with respect to θ. The equation z = √(4a^2 - ρ^2) is explained as part of the boundary of a sphere with equation x^2 + y^2 + z^2 = 4a^2. The conversation ends with a question about the derivation of ρ^2 + z^2 = 4a^2 from ρ = √(x
s3a

## Homework Statement

The problem and solution are included.

## Homework Equations

Double integration.

## The Attempt at a Solution

Firstly, I'd like to mention that the additional ρ under the square root is there accidentally and that it should be outside of the square root such that it forms the ending ρ dρ dθ.

I'm stuck at the sqrt(4a^2 - ρ^2) ρ dρ step. I have a feeling that I need to use trigonometry with ρ = 2acosθ or ρ = 2asinθ but firstly, which of the two would I choose and secondly given that I then have to integrate with respect to θ, must I replace ρ with a variable other than θ? Why or why not?

Any help would be greatly appreciated!

#### Attachments

• Problem.jpg
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I did it with $\rho=2a \sin t$.

Thanks for saying but I also just realized that I don't know why z = √(4a^2 - ρ^2). Could you explain that to me please?

The "sphere of radius 2a", which forms part of the boundary, has equation $x^2+ y^2+ z^2= 4a^2$. In polar or cylindrical coordinates (not spherical coordinates), $\rho=\sqrt{x^2+ y^2}$ so that equation becomes $\rho^2+ z^2= 4a^2$ so that $z^2= 4a^2- \rho^2$ and $z= \pm\sqrt{4a^2- \rho^2}$.

How did you go from ρ = √(x^2 + y^2) to ρ^2 + z^2 = 4a^2?

Change $\rho=\sqrt{x^2+y^2}$ in $x^2+y^2+z^2=4a^2$

## 1. What is double integration and why is the square root step important?

Double integration is a mathematical technique used to solve for the area under a curve in two dimensions. The square root step is important because it allows us to convert the variable of integration into a form that is easily solvable.

## 2. How do I know when to use double integration?

Double integration is typically used when you need to find the area under a curve in two dimensions. It is also used to solve for volume in three dimensions. If you have a problem that involves a curve or shape, double integration may be a viable solution.

## 3. What is the process for performing double integration?

The process for double integration involves breaking down the problem into smaller parts, determining the limits of integration, setting up the double integral expression, solving for the inner integral, and then solving for the outer integral.

## 4. How do I handle the square root step in double integration?

To handle the square root step in double integration, you can use substitution to change the variable of integration into a form that is easily solvable. This may involve using trigonometric identities or other mathematical techniques.

## 5. Can you provide an example of a problem involving double integration and the square root step?

Sure, here is an example: Find the area of the region enclosed by the curves y = x^2 and y = √x. The square root step comes into play when we set up the double integral expression as ∫∫√x - x^2 dydx. We can use substitution by letting u = √x to simplify the problem and solve for the area.

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