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📝 Changing order of integration triple integrals (14 MCQs)

📖 From Calculus • 15. Multiple Integrals Calculus • 14 questions available

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What is Changing order of integration triple integrals?

Definition:
We can change the order of integration (e.g., from $dz \, dy \, dx$ to $dx \, dz \, dy$ ) by re-describing the region's projections and bounds.

Example:
If integrating in $z$ first is difficult, we might switch to integrating in $x$ first if the bounds are simpler in that direction.

Reason:
Changing order can simplify evaluation when one order leads to difficult antiderivatives or complex limits, offering computational flexibility.

Easy

Medium

Hard

📝 All Changing order of integration triple integrals MCQs

Q1. Which transformation correctly changes the order of integration for $I=\int_0^1\int_x^1 f(x,y)\,dy\,dx$ ?

I=\int_0^1\int_0^y f(x,y)\,dx\,dy

✅

I=\int_0^1\int_y^1 f(x,y)\,dx\,dy

I=\int_0^1\int_0^x f(x,y)\,dx\,dy

I=\int_0^1\int_x^1 f(x,y)\,dx\,dy

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The original limits describe the triangular region $0\le x\le1$ and $x\le y\le1$ . For a fixed $y$ , the horizontal slice runs from $x=0$ to $x=y$ , while $y$ ranges from $0$ to $1$ . Therefore the reversed order is $\int_0^1\int_0^y f(x,y)\,dx\,dy$ .

Q2. When reversing $I=\int_0^2\int_{x/2}^1 f(x,y)\,dy\,dx$ , why does the new description require $0\le y\le1$ and different bounds for $x$ ?

A.Because

y=x/2

becomes

x=2y

, and horizontal slices meet

x=2y

and

x=2

✅

B.Because reversing the order always changes every lower limit to zero

C.Because

x/2

becomes

2/x

after reversal

D.Because the region becomes rectangular when the order is reversed

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The boundary $y=x/2$ can be rewritten as $x=2y$ . The original region is bounded by $x=0$ , $x=2$ , $y=x/2$ , and $y=1$ . Horizontal slices therefore have $x$ from $0$ to $2y$ for $0\le y\le1$ , so the reversed integral is $\int_0^1\int_0^{2y}f(x,y)\,dx\,dy$ .

Q3. A region is described by $0\le x\le4$ and $x^2/4\le y\le4$ . A student claims that reversing the order gives $0\le y\le4$ , $0\le x\le2\sqrt y$ . What is the best assessment?

A.Correct, because

y=x^2/4

gives

x=2\sqrt y

✅

B.Incorrect, because

x

must range from

2\sqrt y

4

C.Incorrect, because the upper boundary is

y=4-x

D.Correct only when

y=4

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The region lies above the parabola $y=x^2/4$ and below $y=4$ , with $x$ between $0$ and $4$ . Solving the curved boundary for $x$ gives $x=2\sqrt y$ . Thus each horizontal slice runs from $x=0$ to $x=2\sqrt y$ , so the student's reversed description is correct.

Q4. Suppose a region is bounded by $y=x$ , $y=2x$ , and $x=3$ . Which feature determines whether the reversed integral must be split into more than one integral?

A.Whether horizontal lines intersect the region with different boundary functions over different

y

-ranges ✅

B.Whether the integrand contains

x

y

C.Whether the original integral has two signs in its limits

D.Whether the numerical value of the integral is positive

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Changing the order means describing the same region with horizontal rather than vertical slices. If a horizontal line encounters different boundaries as $y$ changes, one pair of $x$ -limits cannot describe the whole region. Here the lines $y=x$ and $y=2x$ create different intersections, so the reversed description may require splitting.

Q5. A student wants to reverse the order of $I=\int_0^1\int_{x}^{2-x} f(x,y)\,dy\,dx$ . They argue that the new limits can be obtained by simply swapping $x$ and $y$ everywhere. Why is this reasoning insufficient?

A.The limits describe a region, so the boundary curves must be solved and compared using horizontal slices ✅

B.Swapping variables is never allowed in double integrals

C.The integrand must always be integrated first with respect to

x

D.Only rectangular regions can have their order reversed

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reversing the order is not merely a symbolic exchange of $x$ and $y$ . The limits encode a geometric region. One must identify its boundaries, determine the range of the new outer variable, and find the corresponding horizontal intersections. This prevents missing a boundary or assigning an impossible interval.

Q6. Consider the region $0\le x\le2$ , $x^2\le y\le2x$ . Which reversed description represents the same region?

0\le y\le4,\; y/2\le x\le\sqrt y

✅

0\le y\le2,\; y/2\le x\le\sqrt y

0\le y\le4,\; \sqrt y\le x\le y/2

0\le y\le4,\; 0\le x\le\sqrt y

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The boundaries are $y=x^2$ and $y=2x$ . Their intersection occurs at $x=0,2$ , giving $y=0,4$ . Solving the boundaries for $x$ gives $x=\sqrt y$ and $x=y/2$ . For $0<y<4$ , $y/2\le\sqrt y$ , so the horizontal slice is $y/2\le x\le\sqrt y$ .

Q7. A model gives $Q=\int_0^1\int_{x^2}^{x} (x+y)\,dy\,dx$ . If $Q$ represents accumulated output over a region, which reversed-order setup preserves the same region and integrand?

\int_0^1\int_y^{\sqrt y}(x+y)\,dx\,dy

✅

\int_0^1\int_{\sqrt y}^{y}(x+y)\,dx\,dy

\int_0^1\int_0^y(x+y)\,dx\,dy

\int_0^1\int_y^1(x+y)\,dx\,dy

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The original region satisfies $0\le x\le1$ , $x^2\le y\le x$ . Rewriting the boundaries gives $x=\sqrt y$ and $x=y$ . Since $0\le y\le1$ and $y\le\sqrt y$ , each horizontal slice runs from $x=y$ to $x=\sqrt y$ . Hence option A preserves the same accumulated quantity.

Q8. A heat distribution is modeled by $H=\int_0^2\int_{x^2/2}^{2} h(x,y)\,dy\,dx$ . Which strategy is most efficient if $h(x,y)$ becomes much simpler when integrated with respect to $x$ ?

A.Reverse the order after describing the region horizontally ✅

B.Change

h(x,y)

h(y,x)

before integrating

C.Keep the order fixed because changing order changes the physical region

D.Replace the curved boundary by its average value

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The purpose of reversing the order is not merely algebraic; it can make the inner integration substantially easier. Here the region is bounded below by $y=x^2/2$ and above by $y=2$ . Solving the curved boundary gives $x=\sqrt{2y}$ , so horizontal slices can be used to place $x$ inside the inner integral.

Q9. A numerical calculation is difficult for $I=\int_0^1\int_{x}^{\sqrt{x}} e^{y^2}\,dy\,dx$ . Which observation provides the strongest reason to reverse the order?

A.The inner antiderivative of

e^{y^2}

is not elementary, while integrating with respect to

x

may be straightforward ✅

B.The value of every double integral is zero after reversing order

C.The function

e^{y^2}

becomes

e^{x^2}

automatically

D.Reversing order always eliminates one integration

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The expression $e^{y^2}$ has no elementary antiderivative with respect to $y$ , making the original inner integral inconvenient. The region is $0\le x\le1$ , $x\le y\le\sqrt x$ . Reversing the order allows $x$ to be expressed using $y$ , potentially producing an elementary inner integral and simplifying the calculation.

Q10. A student reverses $\int_0^1\int_{x^2}^{x} f(x,y)\,dy\,dx$ as $\int_0^1\int_y^{y^2} f(x,y)\,dx\,dy$ . Which error has occurred?

A.The student reversed the order but placed the horizontal boundaries in the wrong order ✅

B.The student changed the region from curved to rectangular

C.The student forgot that

y

must range from

0

1

D.The student incorrectly changed the integrand

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original boundaries are $y=x^2$ and $y=x$ . Solving for $x$ gives $x=\sqrt y$ and $x=y$ , not $x=y$ and $x=y^2$ . Also, for $0<y<1$ , $y<\sqrt y$ . Thus the correct inner limits are $y$ to $\sqrt y$ , while the proposed $y$ to $y^2$ reverses the geometry.

Q11. A proposed reversal of a region produces $I=\int_0^4\int_{\sqrt y}^{2} f(x,y)\,dx\,dy$ , but the original region is $0\le x\le2$ , $x^2\le y\le4$ . What is wrong with the proposal?

A.Nothing; it describes exactly the same region ✅

B.The lower

x

-limit should be

0

, not

\sqrt y

C.The upper

y

-limit should be

2

, not

4

D.The lower

x

-limit should be

2\sqrt y

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The original region consists of points above $y=x^2$ and below $y=4$ , with $0\le x\le2$ . For a fixed $y$ , solving $y=x^2$ gives $x=\sqrt y$ , and the region extends from that curve to $x=2$ . Therefore the proposed reversed integral is correct.

Q12. A graph shows a shaded region enclosed by $y=x^2$ , $y=2$ , and the $y$ -axis. Which integral correctly represents its area after reversing the natural vertical slicing?

\int_0^2\int_0^{\sqrt y}1\,dx\,dy

✅

\int_0^2\int_y^{\sqrt y}1\,dx\,dy

\int_0^{\sqrt2}\int_0^{y^2}1\,dy\,dx

\int_0^2\int_{\sqrt y}^{2}1\,dx\,dy

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The shaded region is to the right of the $y$ -axis and to the left of the parabola $y=x^2$ , below $y=2$ . A horizontal line at height $y$ runs from $x=0$ to $x=\sqrt y$ . Since $y$ ranges from $0$ to $2$ , the area is represented by option A.

Q13. Two students evaluate $I=\int_0^1\int_0^{1-x} (x+y)\,dy\,dx$ . Student A keeps the original order; Student B reverses it. Both claim their answers must differ because the inner variables are different. Which conclusion is mathematically justified?

A.If both descriptions represent the same region and the function is integrable, the values must agree ✅

B.Student A must be correct because the original order is always preferred

C.Student B must be correct because reversing order increases accuracy

D.Their values can differ whenever the integrand contains both

x

and

y

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The two iterated integrals can use different orders while representing the same two-dimensional accumulation. The region here is the triangle $x\ge0$ , $y\ge0$ , $x+y\le1$ . Changing the order changes only the slicing procedure, not the region or the integrand, so both valid calculations must produce the same value.

Q14. For $I=\int_0^1\int_x^{2-x} f(x,y)\,dy\,dx$ , a computer algebra system struggles with the inner integral. A student proposes reversing the order without first checking the geometry. What is the most reliable next step?

A.Plot or analyze the boundary lines and determine where horizontal slices change their endpoints ✅

B.Numerically approximate the inner integral and assume the order is irrelevant

C.Replace both boundary lines by

y=1

because they meet there

D.Exchange

x

and

y

in the limits immediately

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The region is bounded by $y=x$ and $y=2-x$ , forming a triangle. When horizontal slices are used, the left and right boundaries can change depending on the height. Identifying their intersection and the resulting $y$ -ranges first ensures that the reversed integral accurately covers the entire region without omission or duplication.

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