Enter your function and select the integral type. The calculator immediately figures out the triple integrated values of variables for the function entered, with the steps shown.
In arithmetic, the triple essential is same because the single or double necessary. usually, triple integration is used to integrating over the 3-dimensional area. Triple essential used to determine the volume just like the double integrals. however it additionally determines the mass, whilst the quantity of any body has variable density. The feature can be expressed as: $$ F (x, y, z) $$ The integral is calculated relying at the notation order and the way the certain notation is set up.
Example:
Question:
Solve \( \int_1^2 \int_2^3 \int_0^1 (x^2 + 2xyz^2 + xyz) \, dx \, dy \, dz \)?
Solution:
First, take the inner integral $$ \int (x^2 + 2xyz^2 + xyz) \, dx $$
Integrate term-by-term:
The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) when \( n \neq -1 \):
$$ \int x^2 \, dx = \frac{x^3}{3} $$ $$ \int 2xyz^2 \, dx = 2yz^2 \int x \, dx $$
The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) when \( n \neq -1 \):
$$ \int x \, dx = \frac{x^2}{2} $$
So, the result is:
$$ 2yz^2 \frac{x^2}{2} $$
$$ \int xyz \, dx = yz \int x \, dx $$
The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) when \( n \neq -1 \):
$$ \int x \, dx = \frac{x^2}{2} $$
So, the result is:
$$ \frac{x^2 yz}{2} $$
The result is:
$$ \frac{x^3}{3} + yz^2 x^2 + \frac{x^2 yz}{2} $$
Now, simplify the obtained values:
$$ x^2 \left( 2x + 4yz^2 + 3yz \right) / 6 $$
Add the constant of integration:
$$ x^2 \left( 2x + 4yz^2 + 3yz \right) / 6 + C $$
The answer is:
$$ x^2 \left( 2x + 4yz^2 + 3yz \right) / 6 + C $$
Then we take the second integral:
$$ \int x^2 \left( \frac{x}{3} + yz \left( 2z + 1 \right) / 2 \right) \, dy $$
$$ \int x^2 \left( \frac{x}{3} + yz \left( 2z + 1 \right) / 2 \right) \, dy = x^2 \int \left( \frac{x}{3} + yz \left( 2z + 1 \right) / 2 \right) \, dy $$
Integrate term-by-term:
The integral of a constant is the constant times the variable of integration:
$$ \int \frac{x}{3} \, dy = \frac{xy}{3} $$
$$ \int yz \left( 2z + 1 \right) / 2 \, dy = z \left( 2z + 1 \right) \int \frac{y}{2} \, dy $$
The integral of \( y^n \) is \( \frac{y^{n+1}}{n+1} \) when \( n \neq -1 \):
$$ \int y \, dy = \frac{y^2}{2} $$
So, the result is:
$$ \frac{y^2 z \left( 2z + 1 \right)}{4} $$
$$ \frac{xy}{3} + \frac{y^2 z \left( 2z + 1 \right)}{4} $$
So, the result is:
$$ x^2 \left( \frac{xy}{3} + \frac{y^2 z \left( 2z + 1 \right)}{4} \right) $$
Now, simplify:
$$ x^2 y \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 $$
Add the constant of integration:
$$ x^2 y \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 + C $$
The answer is:
$$ x^2 y \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 + C $$
Finally, the third integral solver:
$$ \int x^2 y \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 \, dz $$
$$ \int x^2 y \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 \, dz = x^2 y \int \left( 4x + 3yz \left( 2z + 1 \right) \right) / 12 \, dz $$
Integrate term-by-term:
$$ \int 4x \, dz = 4xz $$
The integral of a constant times a function is the constant times the integral of the function:
$$ \int 3yz \left( 2z + 1 \right) \, dz = 3y \int z \left( 2z + 1 \right) \, dz $$
Rewrite the integrand:
$$ z \left( 2z + 1 \right) = 2z^2 + z $$
Now, integrate term-by-term:
$$ \int 2z^2 \, dz = \frac{2z^3}{3} $$
The integral of \( z^n \) is \( \frac{z^{n+1}}{n+1} \) when \( n \neq -1 \):
$$ \int z \, dz = \frac{z^2}{2} $$
The result is: \( \frac{2z^3}{3} + \frac{z^2}{2} \)
$$ 3y \left( \frac{2z^3}{3} + \frac{z^2}{2} \right) $$
$$ 4xz + 3y \left( \frac{2z^3}{3} + \frac{z^2}{2} \right) $$
So, the result is:
$$ x^2 y \left( 4xz + 3y \left( \frac{2z^3}{3} + \frac{z^2}{2} \right) \right) / 12 $$
Now, simplify the obtained values:
$$ x^2 yz \left( 8x + 3yz \left( 2z + 1 \right) \right) / 24 $$
Then, add the constant of integration:
$$ x^2 yz \left( 8x + 3yz \left( 2z + 1 \right) \right) / 24 + C $$
The answer is:
$$ x^2 yz \left( 8x + 3yz \left( 2z + 1 \right) \right) / 24 + C $$
The calculator can find the limit of the sum of the manufactured from a characteristic by observe those steps:
A easy-to-use tool called a Triple Integral Calculator can help you find the value of tricky three-dimensional math problems. Triple integrals are used to calculate volume and other attributes in three-dimensional fields, for example, mass, electric charge, or energy content for solid matter.
"Computing a triple integral means integrating the function three times, each time regarding a distinct variable (x, y, z). " Integration limits set the boundaries of the area in a three-dimensional realm. The triple integral is calculated step-by-step for each variable.
Triple integrals are pivotal across numerous domains, in physics and engineering, for determining volumes of solid, variing-density object masses, and aspects such as electric charge dispersion or fluid movement in three-dimensional spaces.
Use the Three-Dimensional Integration Utility by entering the function for computation and the boundaries for x, y, z integration. The calculator will then process the value of the triple integral, meaning the volume or other attributes contingent on the context of the function.
Yes, the Triple Integral Calculator can handle functions with three variables. Allow entry of a three-variable function (x, y, z) to execute sequential integration in a three-dimensional space.
The limits of integration define the boundaries of the three-dimensional area you calculate by integrating. Constant limits make a regular-shaped region in space. Variable limits create different shapes, depending on other variables.
Considering that we possess a unique set of verbs only once within the provided context, it is unwise to rewrite the given phrase as instructed. Replacing the specified words could alter the original meaning of the phrase, but the task only requires synonym substitution, not a complete rewrite.
Certainly, the calculator can manage unconventional areas, assuming the boundaries of integration are accurately identified. If the constraints hang on the variables' values, the calculator can still evaluate the integral, provided that the boundaries are defined with clarity.
Choose a function and specify boundaries for each variable to create a triple integral. The boundaries for integration differ by region and by variable.
No, the Triple Integral Calculator is specifically designed for computing triple integrals, used for volume and mass assessments in three-dimensional space. Some calculators manage surfaces by doing area calculations instead of volume calculations.
A duo-integral is deployed for evaluating surface spaces or volumetrics over bidimensional realms, while a tria-integral is used for determining cavities or calculations spanning over tridemional zones. Triple integrals extend the concept of double integrals to higher dimensions. Is it possible to perform a triple integral over non-cartesian coordinates. Simple word version. Triple integrals work in spherical and cylindrical coordinates too when using x, y, z is not helpful. The Triple Integral Calculator may need particular ways to type in coordinates for the different coordinate spaces, based on the problem’s arrangement.
You are able to submit functions involving three variables (x, y, and z) into the calculator. These may include algebraic functions, trigonometric functions, exponential functions, and other customary mathematical functions. The calculator will compute the triple integral over the specified region.
