The triple integral as the words states is the three integrals applied to a given function. It is usually used to find the area in three dimensional figure. The first integral includes limit values of x, the second integral includes limit values of y and third includes limit values of z respectively.

The triple integral is of the form

Triple Integral Calculator integrates thrice the given function if the function, order and limits are entered in the given space.

## Triple Integral Problems

Steps to solve the Triple Integral:
1. Read the given problem and observe Whether the given function is triple integral.
2. Observe the limits and integrate each variable seperately and apply the limits accordingly.
3. After integrating all the variables from inner to outer integral simplify it and get the answer.

Below are given some problems based on triple integral which may be useful to you.

### Solved Examples

Question 1: Evaluate the triple integral for $\int_{0}^{1}$ $\int_{1}^{2}$ $\int_{2}^{3}$ xyz dx dy dz
Solution:

Given Integral is
$\int_{2}^{3}$ $\int_{1}^{2}$ $\int_{0}^{1}$ xyz dx dy dz = $\int_{2}^{3}$ $\int_{1}^{2}$ $\frac{x^{2}}{2}$ $|_{0}^{1}$ yz dy dz

= $\int_{2}^{3}$ $\int_{1}^{2}$ $\frac{1 - 0}{2}$ yz dy dz

= $\frac{1}{2}$ $\int_{2}^{3}$ $\frac{y^{2}}{2}$ $|_{1}^{2}$ z dz

= $\frac{1}{2}$ $\int_{2}^{3}$ $\frac{2 - 1}{2}$ z dz

= $\frac{1}{4}$ $\int_{2}^{3}$ z dz

= $\frac{1}{4}$ $\frac{z^{2}}{2}$ $|_{2}^{3}$

= $\frac{1}{4}$ $\frac{3^{2} - 2^{2}}{2}$

= $\frac{1}{4}$ $\frac{5}{2}$

=  $\frac{5}{8}$.

Question 2: Evaluate the triple integral for $\int_{0}^{1}$ $\int_{1}^{2}$ $\int_{2}^{3}$ (x + yz) dx dy dz
Solution:

Given Integral is
$\int_{2}^{3}$ $\int_{1}^{2}$ $\int_{0}^{1}$ (x + yz) dx dy dz = $\int_{0}^{1}$ $\int_{1}^{2}$ (xz + $\frac{yz^{2}}{2}$ $|_{2}^{3}$ ) dx dy
= $\int_{0}^{1}$ $\int_{1}^{2}$ (3x + $\frac{3^{2} y}{2}$ - 2x - 2y ) dx dy
= $\int_{0}^{1}$ $\int_{1}^{2}$ $\frac{x + 5y}{2}$ $|_{1}^{2}$ dx dy
= $\int_{0}^{1}$ (x + $\frac{5y}{2}$) $|_{2}^{3}$ dx dy
= $\int_{0}^{1}$ (xy + $\frac{5y^{2}}{4}$) $|_{1}^{2}$ dx
= $\int_{0}^{1}$ (2x + 5 - x - $\frac{5}{4}$) $|_{1}^{2}$
= $\int_{0}^{1}$ (x + $\frac{15}{4}$)
= ($\frac{x^{2}}{2}$ + $\frac{15}{4}$ x $|_{0}^{1}$
= $\frac{1}{2}$ + $\frac{15}{4}$ - 0 + 0
= $\frac{17}{4}$.
Even if you integrate the components in reverse order you get the same answer.

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