Question

Write the integral \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) in the five other possible orders of integration.

Short answer

Question: Rewrite the given triple integral in all six possible orderings and identify any identical orderings. The original integral is: \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\). Answer: The six possible orderings are: 1. \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) 2. \(\int_{0}^{2} \int_{0}^{1-y} \int_{0}^{1} d y d z d x\) 3. \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-z} d x d y d z\) Orderings 1, 4, and 6 are the same, and orderings 3 and 5 are the same; therefore, there are three distinct orderings.

Step-by-step solution

Identify the region of integration

Plot the points \((x,y,z)\) that satisfy the conditions given by the limits of integration: - \(0 \leq x \leq 2\) - \(0 \leq y \leq 1\) - \(0 \leq z \leq 1-y\) We can see that the region is a triangular prism with vertices at \((0,0,0), (2,0,0), (2,1,0), (0,1,0), (0,1,1),\) and \((2,1,1)\). Now that we know the region's shape, we can find the other possible orderings:

Rewrite the integral in all six possible orderings

The six possible permutations of \(d x d y d z\) are: 1. \(d x d y d z\) 2. \(d x d z d y\) 3. \(d y d x d z\) 4. \(d y d z d x\) 5. \(d z d x d y\) 6. \(d z d y d x\) We already have the first ordering, so let's find the other five: 2. \(d x d z d y\): Note that \(y\) has to be the inner integral; the lower limit fixed at \(0\) and the upper limit as \(1-z\). The other two limits will be the same as the given integral. So, the integral is: \(\int_{0}^{2} \int_{0}^{1-y} \int_{0}^{1} d y d z d x\) 3. \(d y d x d z\): In this case, \(z\) has to be the outer integral; the two limits will be the same as the given integral. For the inner limits, \(x\) varies from \(0\) to \(2\) and \(y\) varies from \(0\) to \(1-z\): \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-z} d x d y d z\) 4. \(d y d z d x\): In this ordering, \(x\) has to be the outer integral. Both limits will be the same as in the given integral. For the other two limits, \(y\) will go from \(0\) to \(1\), and \(z\) will go from \(0\) to \(1 - y\): \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) 5. \(d z d x d y\): The outer integral must be \(y\), with both limits fixed at \(0\) and \(1\). The inner limits will have \(x\) going from \(0\) to \(2\) and \(z\) going from \(0\) to \(1-y\): \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-y} d x d z d y\) 6. \(d z d y d x\): The outer integral in this case is \(x\), with both limits \(0\) and \(2\). The inner limits will have \(y\) going from \(0\) to \(1\) and \(z\) going from \(0\) to \(1-y\): \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d y d z d x\) So, the given integral can be rewritten in all six possible orderings like this: 1. \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) 2. \(\int_{0}^{2} \int_{0}^{1-y} \int_{0}^{1} d y d z d x\) 3. \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-z} d x d y d z\) 4. \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) (This is the same as the first ordering.) 5. \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-y} d x d z d y\) (This is the same as ordering 3.) 6. \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d y d z d x\) (This is the same as the first ordering.) We can see that orderings 1, 4, and 6 are the same, and orderings 3 and 5 are the same. That gives us three distinct orderings: 1. \(\int_{0}^{2} \int_{0}^{1} \int_{0}^{1-y} d z d y d x\) 2. \(\int_{0}^{2} \int_{0}^{1-y} \int_{0}^{1} d y d z d x\) 3. \(\int_{0}^{1} \int_{0}^{2} \int_{0}^{1-z} d x d y d z\)