Changing Integration Order - Multivariable Calculus

The sum of integrals $\int_0^1 \int_0^{2y} f(x,y) dx dy + \int_1^2 \int_0^{4-2y} f(x,y) dx dy$ represents the integration of a function $f(x,y)$ over a planar region $R$ . Which single iterated integral also represents the integration of $f(x,y)$ over $R$ ?
Answer: $\int_0^2 \int_{x/2}^{2-x/2} f(x,y) dy dx$ The first integral is over the region $R_1$ defined by $0 \le y \le 1$ and $0 \le x \le 2y$ . This is a triangle with vertices $(0,0)$ , $(2,1)$ , and $(0,1)$ . The second integral is over the region $R_2$ defined by $1 \le y \le 2$ and $0 \le x \le 4-2y$ . This is a triangle with vertices $(0,1)$ , $(2,1)$ , and $(0,2)$ . The total region $R = R_1 \cup R_2$ is a triangle with vertices $(0,0)$ , $(2,1)$ , and $(0,2)$ . To express this as a single integral with the order reversed ( $dy dx$ ), we need to find the bounds for $y$ in terms of $x$ . The overall range for $x$ is from $0$ to $2$ . The lower boundary of the region is the line segment from $(0,0)$ to $(2,1)$ , which has the equation $y=x/2$ . The upper boundary is the line segment from $(2,1)$ to $(0,2)$ , which has the equation $y-2 = \frac{1-2}{2-0}(x-0)$ , or $y = -x/2 + 2$ . Thus, for a fixed $x$ from $0$ to $2$ , $y$ ranges from $x/2$ to $2-x/2$ . The equivalent single integral is $\int_0^2 \int_{x/2}^{2-x/2} f(x,y) dy dx$ .
Let $R$ be the region in the first quadrant bounded by the parabola $y=x^2$ , the line $y=4$ , and the y-axis. The integral of a function $f(x,y)$ over $R$ can be written as $I = \iint_R f(x,y) dA$ . If one were to set up this integral with $y$ as the inner variable of integration ( $dy dx$ ), and then reverse the order, what would be the resulting integral?
Answer: $\int_0^4 \int_0^{\sqrt{y}} f(x,y) dx dy$ When you encounter double integral order-reversal problems, you need to carefully analyze the region of integration and express the same region using different limits. First, let's establish the region R. In the first quadrant, you have the parabola $y = x^2$ , the horizontal line $y = 4$ , and the y-axis ( $x = 0$ ). These intersect where $x^2 = 4$ , so $x = 2$ (taking the positive value since we're in the first quadrant). The region R is bounded by $0 ≤ x ≤ 2$ and $x^2 ≤ y ≤ 4$ . Setting up with $dy dx$ order first: $\int_0^2 \int_{x^2}^4 f(x,y) dy dx$ To reverse the order to $dx dy$ , you need to think about how x varies for each fixed y-value. Since y ranges from 0 to 4, and for each y-value, x ranges from the y-axis ( $x = 0$ ) to the parabola, you need to solve $y = x^2$ for x, giving $x = \sqrt{y}$ . Therefore: $\int_0^4 \int_0^{\sqrt{y}} f(x,y) dx dy$ This matches answer choice B. Answer A gives the original integral before order reversal. Answer C incorrectly uses $y^2$ instead of $\sqrt{y}$ as the upper x-limit, confusing the relationship between x and y on the parabola. Answer D has the wrong y-limits (0 to 2 instead of 0 to 4), mixing up the x-range with the y-range. Strategy tip: Always sketch the region first, then carefully trace how each variable varies while the other is held fixed. The key is correctly inverting the boundary equation.
For the region bounded by the curves $y = x^2$ , $y = 2x$ , and $x = 0$ , the integral $\int \int_R xy \, dA$ can be written as $\int_0^2 \int_{x^2}^{2x} xy \, dy \, dx$ . What is the equivalent integral with the order of integration reversed?
Answer: $\int_0^1 \int_{y/2}^{\sqrt{y}} xy \, dx \, dy + \int_1^4 \int_{y/2}^{2} xy \, dx \, dy$ The curves y = x² and y = 2x intersect when x² = 2x, so x = 0 or x = 2. The region is bounded by x = 0, y = x², and y = 2x for 0 ≤ x ≤ 2. To reverse the order, we need to determine the y-range and corresponding x-bounds. The maximum y-value occurs at the intersection point (2, 4). For 0 ≤ y ≤ 1, the region extends from the line y = 2x (giving x = y/2) to the parabola y = x² (giving x = √y). For 1 ≤ y ≤ 4, the region extends from x = y/2 to x = 2 (the right boundary). The split at y = 1 occurs because that's where the nature of the right boundary changes from the parabola y = x² to the vertical line x = 2.
A solid region is described by the inequalities $0 \le z \le y \le x \le 1$ . The integral of a function $f(x,y,z)$ over this region is given by $\int_0^1 \int_0^x \int_0^y f(x,y,z) dz dy dx$ . Which of the following integrals is equivalent?
Answer: $\int_0^1 \int_z^1 \int_y^1 f(x,y,z) dx dy dz$ The region is defined by $0 \le z \le y \le x \le 1$ . To change the order of integration to $dx dy dz$ , we determine the bounds from the outermost integral inwards. The variable $z$ is bounded by $0$ and can reach a maximum value of $1$ (when $z=y=x=1$ ), so $0 \le z \le 1$ . For a fixed $z$ , the variable $y$ is bounded below by $z$ and can reach a maximum of $1$ , so $z \le y \le 1$ . For fixed $y$ and $z$ , the variable $x$ is bounded below by $y$ and above by $1$ , so $y \le x \le 1$ . Combining these gives the integral $\int_0^1 \int_z^1 \int_y^1 f(x,y,z) dx dy dz$ .
The integral $\int_0^1 \int_{y^2}^{2y} \int_0^{x+y} f(x,y,z) \, dz \, dx \, dy$ represents integration over a region in three-dimensional space. When the order is changed to dx dz dy, which expression gives the correct bounds?
Answer: $\int_0^1 \int_0^{3} \int_{\max(y^2, z-y)}^{\min(2y, z-y)} f(x,y,z) \, dx \, dz \, dy$ From the original bounds: 0 ≤ y ≤ 1, y² ≤ x ≤ 2y, and 0 ≤ z ≤ x+y. The maximum value of z occurs when x = 2y and y = 1, giving z_max = 2(1) + 1 = 3. So z ranges from 0 to 3. For fixed y and z, we need y² ≤ x ≤ 2y and 0 ≤ z ≤ x+y. The constraint z ≤ x+y gives x ≥ z-y. Also, since z ≥ 0, we need z ≥ 0. The bounds for x become max(y², z-y) ≤ x ≤ min(2y, no upper constraint from z since z ≤ x+y is automatically satisfied when x ≥ z-y). Wait, let me reconsider: if z ≤ x+y, then the maximum x for given y,z should incorporate this. Actually, the original constraint 0 ≤ z ≤ x+y means z-y ≤ x, so x ≥ z-y. The upper bound for x remains 2y from the original constraint. So max(y², z-y) ≤ x ≤ 2y, but we also need this to be a valid interval.
Consider the integral $I = \int_0^{1/\sqrt{2}} \int_y^{\sqrt{1-y^2}} f(x,y) dx dy$ . After reversing the order of integration, the integral is expressed as a sum of two integrals. What is this sum?
Answer: $\int_0^{1/\sqrt{2}} \int_0^x f(x,y) dy dx + \int_{1/\sqrt{2}}^1 \int_0^{\sqrt{1-x^2}} f(x,y) dy dx$ The region of integration is given by $0 \le y \le 1/\sqrt{2}$ and $y \le x \le \sqrt{1-y^2}$ . This is a sector of the unit circle in the first quadrant. The left boundary is the line $x=y$ (or $y=x$ ), and the right boundary is the circle $x=\sqrt{1-y^2}$ (or $x^2+y^2=1$ ). The region spans from the line $y=x$ (a 45-degree line) down to the x-axis. To reverse the order of integration to $dy dx$ , we must split the region at the x-coordinate where the top boundary changes. The line $y=x$ and the circle $y=\sqrt{1-x^2}$ intersect when $x = 1/\sqrt{2}$ . For $x$ between $0$ and $1/\sqrt{2}$ , the region is bounded above by $y=x$ and below by $y=0$ . For $x$ between $1/\sqrt{2}$ and $1$ , the region is bounded above by $y=\sqrt{1-x^2}$ and below by $y=0$ . This gives the sum of two integrals: $\int_0^{1/\sqrt{2}} \int_0^x f(x,y) dy dx + \int_{1/\sqrt{2}}^1 \int_0^{\sqrt{1-x^2}} f(x,y) dy dx$ .
Evaluate the iterated integral $\int_0^8 \int_{\sqrt[3]{y}}^2 \cos(x^4) dx dy$ by changing the order of integration.
Answer: $\frac{1}{4}\sin(16)$ When you encounter an iterated integral that's difficult to evaluate directly, changing the order of integration is often the key to solving it. The integrand $\cos(x^4)$ has no elementary antiderivative with respect to $x$ , so you must switch the order. First, sketch the region of integration. The limits tell you that $y$ goes from 0 to 8, and for each $y$ , $x$ goes from $\sqrt[3]{y}$ to 2. The curve $x = \sqrt[3]{y}$ means $y = x^3$ . So your region is bounded by $y = x^3$ , $y = 8$ , and $x = 2$ . To reverse the order, determine the new limits. Since $x$ ranges from 0 to 2, and for each $x$ , $y$ goes from $x^3$ to 8, the integral becomes: $\int_0^2 \int_{x^3}^8 \cos(x^4) dy dx$ Now you can integrate with respect to $y$ first: $\int_{x^3}^8 \cos(x^4) dy = \cos(x^4) \cdot y \Big|_{x^3}^8 = \cos(x^4)(8 - x^3)$ This gives you: $\int_0^2 \cos(x^4)(8 - x^3) dx = \int_0^2 8\cos(x^4) dx - \int_0^2 x^3\cos(x^4) dx$ For the second integral, use substitution $u = x^4$ , so $du = 4x^3 dx$ . This yields $\frac{1}{4}\sin(16)$ for the entire expression. Choice A gives $4\sin(16)$ (missing the factor of $\frac{1}{4}$ ), choice B gives $\sin(16)$ (computational error), and choice C involves cosine instead of sine, suggesting confusion with the antiderivative. Always sketch the region first when changing integration order—visualization prevents limit errors.
Which of the following iterated integrals is equivalent to the expression $I = \int_0^1 \int_{x^2}^{\sqrt{x}} e^{y^{3/2}} dy dx$ ?
Answer: $\int_0^1 \int_{y^2}^{\sqrt{y}} e^{y^{3/2}} dx dy$ The region of integration is defined by $0 \le x \le 1$ and $x^2 \le y \le \sqrt{x}$ . To change the order of integration, we first analyze the region. The curves $y=x^2$ and $y=\sqrt{x}$ intersect at $(0,0)$ and $(1,1)$ . For $x \in [0,1]$ , we have $x^2 \le \sqrt{x}$ . If we integrate with respect to $x$ first, we need to express the bounds of $x$ in terms of $y$ . The range of $y$ over this region is from $0$ to $1$ . The curve $y=\sqrt{x}$ can be written as $x=y^2$ , and $y=x^2$ can be written as $x=\sqrt{y}$ . For a fixed $y$ between $0$ and $1$ , $x$ goes from the left curve $x=y^2$ to the right curve $x=\sqrt{y}$ . Therefore, the equivalent integral is $\int_0^1 \int_{y^2}^{\sqrt{y}} e^{y^{3/2}} dx dy$ . The change of order is useful here because the original inner integral is not an elementary function, but the new inner integral, $\int e^{y^{3/2}} dx = x e^{y^{3/2}}$ , is easily computed.
The sum of two iterated integrals is given by $I = \int_0^1 \int_{-\sqrt{y}}^0 f(x,y) dx dy + \int_1^2 \int_{-\sqrt{2-y}}^0 f(x,y) dx dy$ . When $I$ is rewritten as a single iterated integral with the order of integration reversed ( $dy dx$ ), what is the equation for the upper bound of the inner integral?
Answer: $y = 2-x^2$ When you see a problem asking to reverse the order of integration in iterated integrals, you need to sketch the region of integration to understand its boundaries in the new coordinate system. Let's analyze each integral separately. The first integral $\int_0^1 \int_{-\sqrt{y}}^0 f(x,y) dx dy$ describes a region where $y$ goes from 0 to 1, and for each $y$ , $x$ goes from $-\sqrt{y}$ to 0. This gives us the curve $x = -\sqrt{y}$ , or equivalently $y = x^2$ (since $x \leq 0$ ). The second integral $\int_1^2 \int_{-\sqrt{2-y}}^0 f(x,y) dx dy$ has $y$ from 1 to 2, with $x$ from $-\sqrt{2-y}$ to 0. This gives us $x = -\sqrt{2-y}$ , which rearranges to $y = 2-x^2$ . When you sketch these regions, you'll see they form a single connected region bounded on the right by $x = 0$ , below by $y = x^2$ , and above by $y = 2-x^2$ . The $x$ -values range from some negative value to 0. For the reversed integral $\int \int f(x,y) dy dx$ , the upper bound of the inner integral (the $dy$ integral) is $y = 2-x^2$ , making (D) correct. The wrong answers represent other boundaries: (A) $y = 2-x$ is a line, not our parabolic boundary; (B) $y = x^2$ is actually the lower boundary; (C) $y = 1$ is just the horizontal line where the two original regions meet. Strategy tip: Always sketch the region first when reversing integration order—visual understanding prevents boundary confusion.
Consider the integral $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(x,y) \, dy \, dx$ . If this integral is rewritten in terms of $\int \int g(r,\theta) \, dr \, d\theta$ , what would be the most appropriate choice for the limits and integrand?
Answer: $\int_0^{2\pi} \int_0^1 f(r\cos\theta, r\sin\theta) \cdot r \, dr \, d\theta$ The original integral represents integration over a disk of radius 1 centered at the origin (the region where x² + y² ≤ 1). When converting to polar coordinates, we have x = r cos θ and y = r sin θ, and the area element becomes dx dy = r dr dθ. The region is the full disk, so r ranges from 0 to 1 and θ ranges from 0 to 2π to cover the entire circle. The integrand becomes f(r cos θ, r sin θ) · r, where the factor r comes from the Jacobian of the coordinate transformation. Choice B only covers half the disk (θ from 0 to π), choice C is missing the Jacobian factor r, and choice D uses the correct range for θ but in a less standard form.
Which of the following integrals is equivalent to $\int_0^1 \int_{e^x}^e f(x,y) dy dx$ ?
Answer: $\int_1^e \int_0^{\ln(y)} f(x,y) dx dy$ The region of integration is defined by $0 \le x \le 1$ and $e^x \le y \le e$ . Let's sketch the region. It is bounded by the y-axis ( $x=0$ ), the line $x=1$ , the curve $y=e^x$ , and the line $y=e$ . At $x=0$ , $y=e^0=1$ . At $x=1$ , $y=e^1=e$ . The region is above $y=e^x$ and below $y=e$ . To reverse the order of integration to $dx dy$ , we need to find the range of $y$ first. The minimum value of $y$ is $1$ (at $x=0$ ) and the maximum is $e$ . So, $1 \le y \le e$ . For a fixed value of $y$ in this range, $x$ is bounded on the left by the y-axis ( $x=0$ ) and on the right by the curve $y=e^x$ , which can be written as $x=\ln(y)$ . Thus, the bounds for $x$ are $0 \le x \le \ln(y)$ . The equivalent integral is $\int_1^e \int_0^{\ln(y)} f(x,y) dx dy$ .
Consider the integral $I = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} f(x,y,z) dz dy dx$ . Which integral represents $I$ with the order of integration changed to $dy dx dz$ ?
Answer: $\int_0^1 \int_0^{1-z} \int_0^{1-z-x} f(x,y,z) dy dx dz$ When changing the order of integration in a triple integral, you need to carefully analyze the region of integration and rewrite the bounds accordingly. The key is understanding what geometric region the original integral describes, then expressing that same region with different variable orderings. The original integral $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f(x,y,z) \, dz \, dy \, dx$ describes a region where: - $x$ ranges from 0 to 1 - For each $x$ , $y$ ranges from 0 to $1-x$ - For each $(x,y)$ pair, $z$ ranges from 0 to $1-x-y$ This defines the simplex region where $x + y + z \leq 1$ and all variables are non-negative. To change to order $dy \, dx \, dz$ , we integrate $z$ first (outermost), then $x$ , then $y$ (innermost). For the simplex: - $z$ ranges from 0 to 1 - For each $z$ , we need $x + y \leq 1-z$ , so $x$ ranges from 0 to $1-z$ - For each $(z,x)$ pair, $y$ ranges from 0 to $1-z-x$ This gives us option C: $\int_0^1 \int_0^{1-z} \int_0^{1-z-x} f(x,y,z) \, dy \, dx \, dz$ Option A incorrectly has $y$ in the bound for $x$ , violating the integration order. Option B has incorrect bounds that don't respect the constraint $x + y + z \leq 1$ . Option D has the variables in the wrong order entirely, with $dy$ appearing twice in the differential ordering. Study tip: Always sketch the region when possible, and remember that each variable's bounds can only depend on the variables integrated in outer loops, never inner ones.
Consider the region R bounded by the surfaces $z = x^2 + y^2$ , $z = 8 - x^2 - y^2$ , and $x^2 + y^2 = 4$ . The integral $\int \int \int_R f(x,y,z) \, dV$ can be written as $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+y^2}^{8-x^2-y^2} f(x,y,z) \, dz \, dy \, dx$ . What is the equivalent integral with order dx dy dz?
Answer: $\int_0^4 \int_{-\sqrt{z}}^{\sqrt{z}} \int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}} f(x,y,z) \, dx \, dy \, dz + \int_4^8 \int_{-\sqrt{8-z}}^{\sqrt{8-z}} \int_{-\sqrt{8-z-y^2}}^{\sqrt{8-z-y^2}} f(x,y,z) \, dx \, dy \, dz$ The region is bounded by the paraboloid z = x²+y² (below), the paraboloid z = 8-x²-y² (above), and the cylinder x²+y² = 4. These surfaces intersect when x²+y² = 8-x²-y², giving x²+y² = 4, which matches the cylindrical boundary. The z-values range from 0 (at the origin on the lower paraboloid) to 8 (at the origin on the upper paraboloid), but we're constrained by the cylinder x²+y² ≤ 4. At the cylinder boundary, z ranges from 4 to 4 (the paraboloids intersect). For 0 ≤ z ≤ 4, the lower paraboloid z = x²+y² gives the constraint x²+y² ≤ z, while for 4 ≤ z ≤ 8, the upper paraboloid z = 8-x²-y² gives x²+y² ≤ 8-z. This splits the integral into two parts as shown in choice C.
Reversing the order of integration for $\int_0^1 \int_{x^2}^x \sin(x^2+y^2) dy dx$ results in which integral? You are not required to evaluate the integral.
Answer: $\int_0^1 \int_y^{\sqrt{y}} \sin(x^2+y^2) dx dy$ When you encounter a problem asking you to reverse the order of integration, you need to carefully analyze the region of integration by sketching it or identifying its boundaries, then rewrite the limits to integrate in the opposite order. The original integral $\int_0^1 \int_{x^2}^x \sin(x^2+y^2) dy dx$ integrates first with respect to $y$ , then $x$ . The region is defined by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$ . Since we need $x^2 \leq x$ , this means $0 \leq x \leq 1$ (which matches our outer limits). The region is bounded below by the parabola $y = x^2$ and above by the line $y = x$ . To reverse the order, you integrate first with respect to $x$ , then $y$ . For a fixed $y$ -value, you need to determine the range of $x$ -values in the region. The $y$ -values range from 0 to 1 (the maximum value where the curves intersect). For a given $y$ between 0 and 1, the leftmost $x$ -value occurs where $x = y$ (from the line), and the rightmost where $x^2 = y$ , so $x = \sqrt{y}$ . Therefore, $y \leq x \leq \sqrt{y}$ . This gives us $\int_0^1 \int_y^{\sqrt{y}} \sin(x^2+y^2) dx dy$ , which is choice C. Choice A incorrectly uses the same functional forms as the original bounds. Choice B reverses the $x$ -limits incorrectly. Choice D simply swaps the integral signs without properly changing the limits. Study tip: Always sketch the region of integration when reversing order. Identify the boundary curves, find their intersection points, then carefully determine how each variable ranges when the other is held fixed.

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