The online triple integral calculator finds the triple integrand values of the given function. The cylindrical integral calculator finds the volume of three-dimensional objects with steps.
In Calculus:
“The analog of double integrals for three dimensions is known as a triple integral”.
It is the same concept as the single and the double integration. It is used to determine the volume like double integrals.
The integration of functions for the three variables can be calculated by the below formula that is considered by the triple integrals calculator. These functions of variables can be represented over the three-dimensional space by w=f(x,y,z).
The following formula is used at the time of triple integration calculation.
$$ \int_{z_1}^{z_2} \left( \int_{y_1}^{y_2} \left( \int_{x_1}^{x_2} f(x,y,z)dx \right) dy \right) dz $$
In order to determine the function with the triple iterated integral calculator or to find it manually you can take into service the following points:
In order to solve the integrations of the variable functions by using the triple integral calculator the following example is the best approach that clarifies their concept.
$$ \int\limits_{0}^{3}\int\limits_{3}^{2}\int\limits_{0}^{1} \left(x^{3} + 5 x y z^{2} + x y z\right)\, dx\, dy\, dz $$
Indefinite Integral
$$ =\frac{x^{2} y z \left(6 x^{2} + 10 y z^{2} + 3 y z\right)}{24}+ \mathrm{constant} $$
Definite Integral
$$ = - \frac{501}{8} $$
Integral Steps:
$$ \int \left(x^{3} + 5 x y z^{2} + x y z\right)\, dx $$
$$ \frac{x^{4}}{4} + x^{2} \left(\frac{5 y z^{2}}{2} + \frac{y z}{2}\right) $$
$$ \frac{x^{2} \left(x^{2} + 2 y z \left(5 z + 1\right)\right)}{4} $$
$$ \frac{x^{2} \left(x^{2} + 2 y z \left(5 z + 1\right)\right)}{4}+ \mathrm{constant} $$
$$ \frac{x^{2} \left(x^{2} + 2 y z \left(5 z + 1\right)\right)}{4}+ \mathrm{constant} $$
$$ \int \frac{x^{2} \left(x^{2} + 2 y z \left(5 z + 1\right)\right)}{4}\, dy $$
The integral of a constant times a function is the constant times the integral of the function:
$$ \int \frac{x^{2} \left(x^{2} + 2 y z \left(5 z + 1\right)\right)}{4}\, dy = \frac{x^{2} \int \left(x^{2} + 2 y z \left(5 z + 1\right)\right)\, dy}{4} $$
The integral of a constant is the constant times the variable of integration:
$$ \int x^{2}\, dy = x^{2} y $$
The integral of a constant times a function is the constant times the integral of the function:
$$ \int 2 y z \left(5 z + 1\right)\, dy = 2 z \left(5 z + 1\right) \int y\, dy $$
$$ \frac{y^{n + 1}}{n + 1} $$ when n≠−1:
$$ \int y\, dy = \frac{y^{2}}{2} $$
$$ y^{2} z \left(5 z + 1\right) $$
$$ x^{2} y + y^{2} z \left(5 z + 1\right) $$
So, the result is: $$ \frac{x^{2} \left(x^{2} y + y^{2} z \left(5 z + 1\right)\right)}{4} $$
Now simplify: $$ \frac{x^{2} y \left(x^{2} + y z \left(5 z + 1\right)\right)}{4} $$
$$ \frac{x^{2} y \left(x^{2} + y z \left(5 z + 1\right)\right)}{4}+ \mathrm{constant} $$
The answer is: $$ \frac{x^{2} y \left(x^{2} + y z \left(5 z + 1\right)\right)}{4}+ \mathrm{constant} $$
$$ \int \frac{x^{2} y \left(x^{2} + y z \left(5 z + 1\right)\right)}{4}\, dz $$
$$ \frac{x^{4} y z}{4} + \frac{5 x^{2} y^{2} z^{3}}{12} + \frac{x^{2} y^{2} z^{2}}{8} $$
$$ \frac{x^{2} y z \left(6 x^{2} + 10 y z^{2} + 3 y z\right)}{24} $$
$$ \frac{x^{2} y z \left(6 x^{2} + 10 y z^{2} + 3 y z\right)}{24}+ \mathrm{constant} $$
The answer is: \frac{x^{2} y z \left(6 x^{2} + 10 y z^{2} + 3 y z\right)}{24}+ \mathrm{constant}
It is extremely helpful to use the triple integral solver if you need fast results of analog for double integrals for three dimensions.
Input:
Output:
An online triple integral calculator with steps gives you the final results as follows:
To perform the triple integrals in the cylindrical coordinates to cartesian coordinates some equations are used that are listed below in the table.
|
Circular Cylinder |
Circular cone |
Sphere |
Paraboloid |
Cylindrical |
R = c |
Z = cr |
\( R^2 + z^2 = c^2 \) |
\( Z = cr^2 \) |
Rectangular |
\( X^2 + y^2 = c^2 \) |
\( Z^2 = c^2 (x^2 + y^2) \) |
\( X^2 + y^2 + z^2 = c^2 \) |
\( Z = c(x^2 + y^2) \) |
Over a more general bounded region to compute the triple integral we take into service all three types
E={(x,y,z)|(x,y)∈D,u1(x,y)≤z≤u2(x,y)}.
For two functions z=u1(x,y) and u2(x,y), such that u1(x,y)≤u2(x,y) for all (x,y) in D as shown in the following figure.
From the source Wikipedia: Triple integral
From the source Libretext: Triple Integrals, Triple Integrals over a General Bounded Region, Triple Integrals in Cylindrical Coordinates
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