Answer
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Hint: Here, we will first define the term triple integral. Then we will elaborate the steps that are involved in evaluating the triple integral of an Integral function using a suitable example. Integration is a process of adding the small parts to find the whole. Integration is the inverse of differentiation and hence it is called antiderivative.
Formula Used:
We will use the following formulas:
1. Product rule of exponents: \[{a^m} \times {a^n} = {a^{m + n}}\]
2. Power rule for Exponents: \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
3. Integral Formula: \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]
Complete Step by Step Solution:
Triple integral is a process of Integration in a three dimensional region. Triple integral is used to compute the volume of a three dimensional region. A Triple integral will be in the form \[\int\limits_{{x_0}}^{{x_1}} {\int\limits_{{y_0}}^{{y_1}} {\int\limits_{{z_0}}^{{z_1}} {f\left( V \right)dzdydx} } } \]
First, we will integrate the inner integral by using the basic integration and then substitute the inner limits.
Then, we will integrate the middle integral and at last integrate the outer integral by using the basic integration and then substitute the middle limits.
Thus, the value obtained after evaluating the outer integral and substituting the outer limits will be the final solution of the Triple integral. We will use the triple integral in the following example.
Example: Evaluate: \[\int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } \]
Now, we will integrate the inner integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{xy{z^2}}}{2}} \right]_1^ydydx} } \]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{xy \cdot {y^2}}}{2} - \dfrac{{xy}}{2}} \right]dydx} } \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{x{y^3}}}{2} - \dfrac{{xy}}{2}} \right]dydx} } \]
Now, we will integrate the middle integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\dfrac{{x{y^4}}}{{2 \cdot 4}} - \dfrac{{x{y^2}}}{{2 \cdot 2}}} \right]_{ - 1}^{{x^2}}dx} \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\dfrac{{x{y^4}}}{8} - \dfrac{{x{y^2}}}{4}} \right]_{ - 1}^{{x^2}}dx} \]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{x{{\left( {{x^2}} \right)}^4}}}{8} - \dfrac{{x{{\left( {{x^2}} \right)}^2}}}{4}} \right) - \left( {\dfrac{{x{{\left( { - 1} \right)}^4}}}{8} - \dfrac{{x{{\left( { - 1} \right)}^2}}}{4}} \right)} \right]dx} \]
Power rule for Exponents: \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
Now, by using power rule for exponents, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{x\left( {{x^8}} \right)}}{8} - \dfrac{{x\left( {{x^4}} \right)}}{4}} \right) - \left( {\dfrac{x}{8} - \dfrac{x}{4}} \right)} \right]dx} \]
Product rule of exponents: \[{a^m} \times {a^n} = {a^{m + n}}\]
Now, by using product rule of exponents, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) - \left( {\dfrac{{x - 2x}}{8}} \right)} \right]dx} \]
Subtracting the like terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) - \left( {\dfrac{{ - x}}{8}} \right)} \right]dx} \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) + \dfrac{x}{8}} \right]dx} \]
Now, we will integrate the middle integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{x^{10}}}}{{8 \cdot 10}} - \dfrac{{{x^6}}}{{4 \cdot 6}}} \right) + \dfrac{{{x^2}}}{{8 \cdot 2}}} \right]_0^2\]
Multiplying the terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{x^{10}}}}{{80}} - \dfrac{{{x^6}}}{{24}}} \right) + \dfrac{{{x^2}}}{{16}}} \right]_0^2\]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{2^{10}}}}{{80}} - \dfrac{{{2^6}}}{{24}}} \right) + \dfrac{{{2^2}}}{{16}}} \right]\]
Applying the exponent on terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{1024}}{{80}} - \dfrac{{64}}{{24}}} \right) + \dfrac{4}{{16}}} \right]\]
Simplifying the fractions, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{64}}{5} - \dfrac{8}{3}} \right) + \dfrac{1}{4}} \right]\]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{64}}{5} \times \dfrac{{12}}{{12}} - \dfrac{8}{3} \times \dfrac{{20}}{{20}}} \right) + \dfrac{1}{4} \times \dfrac{{15}}{{15}}} \right]\]
Multiplying the terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{768}}{{60}} - \dfrac{{160}}{{60}}} \right) + \dfrac{{15}}{{60}}} \right]\]
Simplifying the expression, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \dfrac{{623}}{{60}}\]
Thus, the triple integral of \[\int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } \] is \[\dfrac{{623}}{{60}}\].
Therefore, a triple integral can be solved by following the procedures.
Note:
We need to keep in mind that the outer limits have to be constant and cannot depend on any of the variables. The middle limit depends only on the outer integral and does not depend on the inner integral. The inner limit depends only on the outer integral and the middle integral. Whenever we are substituting the limit, then the lower limit Integrand has to be subtracted from the upper limit value of the Integrand.
Formula Used:
We will use the following formulas:
1. Product rule of exponents: \[{a^m} \times {a^n} = {a^{m + n}}\]
2. Power rule for Exponents: \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
3. Integral Formula: \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]
Complete Step by Step Solution:
Triple integral is a process of Integration in a three dimensional region. Triple integral is used to compute the volume of a three dimensional region. A Triple integral will be in the form \[\int\limits_{{x_0}}^{{x_1}} {\int\limits_{{y_0}}^{{y_1}} {\int\limits_{{z_0}}^{{z_1}} {f\left( V \right)dzdydx} } } \]
First, we will integrate the inner integral by using the basic integration and then substitute the inner limits.
Then, we will integrate the middle integral and at last integrate the outer integral by using the basic integration and then substitute the middle limits.
Thus, the value obtained after evaluating the outer integral and substituting the outer limits will be the final solution of the Triple integral. We will use the triple integral in the following example.
Example: Evaluate: \[\int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } \]
Now, we will integrate the inner integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{xy{z^2}}}{2}} \right]_1^ydydx} } \]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{xy \cdot {y^2}}}{2} - \dfrac{{xy}}{2}} \right]dydx} } \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\left[ {\dfrac{{x{y^3}}}{2} - \dfrac{{xy}}{2}} \right]dydx} } \]
Now, we will integrate the middle integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\dfrac{{x{y^4}}}{{2 \cdot 4}} - \dfrac{{x{y^2}}}{{2 \cdot 2}}} \right]_{ - 1}^{{x^2}}dx} \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\dfrac{{x{y^4}}}{8} - \dfrac{{x{y^2}}}{4}} \right]_{ - 1}^{{x^2}}dx} \]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{x{{\left( {{x^2}} \right)}^4}}}{8} - \dfrac{{x{{\left( {{x^2}} \right)}^2}}}{4}} \right) - \left( {\dfrac{{x{{\left( { - 1} \right)}^4}}}{8} - \dfrac{{x{{\left( { - 1} \right)}^2}}}{4}} \right)} \right]dx} \]
Power rule for Exponents: \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]
Now, by using power rule for exponents, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{x\left( {{x^8}} \right)}}{8} - \dfrac{{x\left( {{x^4}} \right)}}{4}} \right) - \left( {\dfrac{x}{8} - \dfrac{x}{4}} \right)} \right]dx} \]
Product rule of exponents: \[{a^m} \times {a^n} = {a^{m + n}}\]
Now, by using product rule of exponents, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) - \left( {\dfrac{{x - 2x}}{8}} \right)} \right]dx} \]
Subtracting the like terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) - \left( {\dfrac{{ - x}}{8}} \right)} \right]dx} \]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \int\limits_0^2 {\left[ {\left( {\dfrac{{{x^9}}}{8} - \dfrac{{{x^5}}}{4}} \right) + \dfrac{x}{8}} \right]dx} \]
Now, we will integrate the middle integral by using the basic integration, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{x^{10}}}}{{8 \cdot 10}} - \dfrac{{{x^6}}}{{4 \cdot 6}}} \right) + \dfrac{{{x^2}}}{{8 \cdot 2}}} \right]_0^2\]
Multiplying the terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{x^{10}}}}{{80}} - \dfrac{{{x^6}}}{{24}}} \right) + \dfrac{{{x^2}}}{{16}}} \right]_0^2\]
Now, we will substitute the limits, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{{2^{10}}}}{{80}} - \dfrac{{{2^6}}}{{24}}} \right) + \dfrac{{{2^2}}}{{16}}} \right]\]
Applying the exponent on terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{1024}}{{80}} - \dfrac{{64}}{{24}}} \right) + \dfrac{4}{{16}}} \right]\]
Simplifying the fractions, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{64}}{5} - \dfrac{8}{3}} \right) + \dfrac{1}{4}} \right]\]
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{64}}{5} \times \dfrac{{12}}{{12}} - \dfrac{8}{3} \times \dfrac{{20}}{{20}}} \right) + \dfrac{1}{4} \times \dfrac{{15}}{{15}}} \right]\]
Multiplying the terms, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \left[ {\left( {\dfrac{{768}}{{60}} - \dfrac{{160}}{{60}}} \right) + \dfrac{{15}}{{60}}} \right]\]
Simplifying the expression, we get
\[ \Rightarrow \int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } = \dfrac{{623}}{{60}}\]
Thus, the triple integral of \[\int\limits_0^2 {\int\limits_{ - 1}^{{x^2}} {\int\limits_1^y {xyzdzdydx} } } \] is \[\dfrac{{623}}{{60}}\].
Therefore, a triple integral can be solved by following the procedures.
Note:
We need to keep in mind that the outer limits have to be constant and cannot depend on any of the variables. The middle limit depends only on the outer integral and does not depend on the inner integral. The inner limit depends only on the outer integral and the middle integral. Whenever we are substituting the limit, then the lower limit Integrand has to be subtracted from the upper limit value of the Integrand.
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