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:
