Double integration calculation

In summary, to calculate the integral of a sphere with radius a using double integrals in polar coordinates, the volume element must be integrated for r between 0 (midpoint) and R (surface), for theta between 0 (north pole) and pi (south pole), and for phi between 0 (starting from eastward meridian) and 2*pi (arriving at same meridian from west). This results in a volume of (4/3)*pi*R^3. If calculating surface area, use the surface element and integrate for r between 0 and R and for theta between 0 and pi, resulting in a surface area of 4*pi*R^2. It is important to note that the inner integral
  • #1
Fys
16
0
I want to calculate the integral of a sphere with radius a using double integrals

I have [tex] z^2+y^2+x^2=a [/tex] and I plug this in my equation and integrate in polar coördinates

Now i have [tex] \int^{2\pi}_{0} \int ^{a}_{0} \sqrt{(a-r^{2})} rdrd\theta [/tex]

But why do I need to multiply my integral with 2 (because i get half the answer I expect).
Can someone explain this please?

thanks guys
 
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  • #2
I assume that it is the volume of the sphere which you wish to calculate.

In polar coordinates, the volume element is dr * (r * dtheta) * {r * sin(theta) * dphi}.
Or r2 *dr * sin(theta) * dtheta * dphi.

This must be integrated for r between 0 (the midpoint) and R (the surface), for theta between 0 (the north pole) and pi (the south pole) and phi between zero (setting out from some meridian eastward) and 2*pi (arriving at the same meridian from the west).

The integrals can be split.
Integrating r2dr between 0 and R, we get (1/3)R3.
Integrating sin(theta) dtheta between 0 and pi we get 1 - (-1) = 2.
Integrating dphi between 0 and 2*pi, we get 2*pi.

Multiplying, we get (4/3)*pi*R3.

If you want the surface of the sphere, you use the surface element (R*dtheta) *{R*sin(theta)*dphi}, just do the theta and phi integration, and get 4*pi*R2.

Looking at your own method, I get the impression that you are only calculating the volume of the northern hemisphere. The inner integral should run fron -a to +a, not from 0 to +a.
 

1. What is double integration calculation?

Double integration calculation is a mathematical technique used to find the area under a two-dimensional curve. It involves integrating a function twice, first with respect to one variable and then with respect to the other variable.

2. What is the difference between single and double integration?

Single integration involves finding the area under a one-dimensional curve, while double integration involves finding the area under a two-dimensional curve. Double integration is essentially integrating twice, whereas single integration is only integrating once.

3. How is double integration used in real-life applications?

Double integration is commonly used in physics and engineering to calculate the work done by a variable force, the center of mass of an object, and the motion of objects in a two-dimensional space. It is also used in economics to calculate the area under supply and demand curves.

4. What are the steps involved in performing a double integration calculation?

The first step is to determine the limits of integration for both variables. Then, the function is integrated with respect to one variable, using the limits as the upper and lower bounds. Next, the resulting function is integrated again with respect to the other variable, using the remaining limits as the upper and lower bounds. The final step is to evaluate the resulting double integral.

5. Are there any limitations to double integration calculation?

Double integration can become complex when the function being integrated is not well-behaved or the limits of integration are not easily determined. It also requires a good understanding of integration techniques and algebraic manipulation. Additionally, double integration may not always provide an accurate answer when dealing with highly irregular or discontinuous functions.

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