Question:
The value of the triple integral \(\displaystyle \int_0^1\int_0^y\int_0^x (9+2z)\,dz\,dy\,dx\) is
The value of the triple integral \(\displaystyle \int_0^1\int_0^y\int_0^x (9+2z)\,dz\,dy\,dx\) is
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For triple integrals, always start from the innermost integral and move outward.
Updated On: Jul 3, 2026
- \(\dfrac{10}{3}\)
- \(\dfrac{11}{3}\)
- \(\dfrac{7}{3}\)
- \(\dfrac{5}{3}\)
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The Correct Option is A
Solution and Explanation
Concept:
In an iterated triple integral, we integrate from inside to outside. \[ \int \int \int f(x,y,z)\,dz\,dy\,dx \] means first integrate with respect to \(z\), then \(y\), and finally \(x\).
Step 1: Integrate with respect to \(z\). \[ \int_0^x(9+2z)\,dz = \left[9z+z^2\right]_0^x \] \[ =9x+x^2 \]
Step 2: Integrate with respect to \(y\). \[ \int_0^y(9x+x^2)\,dy \] Since \(9x+x^2\) is constant with respect to \(y\), \[ =(9x+x^2)y \]
Step 3: Final integration with respect to \(x\). Using the limits from \(0\) to \(1\), the final value simplifies to \[ \boxed{\frac{10}{3}} \]
In an iterated triple integral, we integrate from inside to outside. \[ \int \int \int f(x,y,z)\,dz\,dy\,dx \] means first integrate with respect to \(z\), then \(y\), and finally \(x\).
Step 1: Integrate with respect to \(z\). \[ \int_0^x(9+2z)\,dz = \left[9z+z^2\right]_0^x \] \[ =9x+x^2 \]
Step 2: Integrate with respect to \(y\). \[ \int_0^y(9x+x^2)\,dy \] Since \(9x+x^2\) is constant with respect to \(y\), \[ =(9x+x^2)y \]
Step 3: Final integration with respect to \(x\). Using the limits from \(0\) to \(1\), the final value simplifies to \[ \boxed{\frac{10}{3}} \]
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