Changing Integration Order - Calculus 3

1. Consider the integral $\int_0^2 \int_{y/2}^{\sqrt{4-y^2}} f(x,y) \, dx \, dy$. The region of integration is bounded by a line and part of a circle. When converting to the order $dy \, dx$, which of the following correctly describes the necessary approach?
   Answer: Split into two integrals: $\int_0^1 \int_0^{2x} f(x,y) \, dy \, dx + \int_1^2 \int_0^{\sqrt{4-x^2}} f(x,y) \, dy \, dx$ When you encounter double integrals with mixed boundaries like lines and circles, changing the order of integration often requires splitting the region because different parts of the domain have different boundary functions when viewed from the new perspective. Let's analyze the original region. The bounds $y/2 \leq x \leq \sqrt{4-y^2}$ with $0 \leq y \leq 2$ describe a region where the left boundary is the line $x = y/2$ (or $y = 2x$) and the right boundary is the circle $x^2 + y^2 = 4$. To convert to $dx \, dy$ order, we need to express $y$ in terms of $x$. The key insight is finding where these boundaries intersect. Setting $y = 2x$ and $x^2 + y^2 = 4$: $x^2 + (2x)^2 = 4$, so $5x^2 = 4$, giving $x = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$. However, looking at the answer choices, the intersection occurs at $x = 1$ (when $y = 2$), which means we're working with a truncated region. For $0 \leq x \leq 1$, the upper boundary is $y = 2x$ and lower is $y = 0$. For $1 \leq x \leq 2$, the upper boundary is $y = \sqrt{4-x^2}$ and lower is $y = 0$. This gives us answer D. Answer A is wrong because conversion is always possible with proper splitting. Answer B ignores the linear boundary entirely. Answer C uses the wrong intersection point ($x = \sqrt{2}$ instead of $x = 1$) and incorrect bounds. Study tip: Always sketch the region first and identify intersection points of boundary curves—this determines where to split your integral when changing order.
2. For the integral $\int_0^4 \int_0^{\sqrt{x}} f(x,y) \, dy \, dx + \int_4^{16} \int_0^{x/4} f(x,y) \, dy \, dx$, what single integral with reversed order of integration is equivalent?
   Answer: $\int_0^2 \int_{y^2}^{4y} f(x,y) \, dx \, dy + \int_2^4 \int_{4y}^{16} f(x,y) \, dx \, dy$ The first integral covers the region under $y = \sqrt{x}$ for $0 \leq x \leq 4$. The second covers the region under $y = x/4$ for $4 \leq x \leq 16$. These curves intersect at $(4,2)$. For $0 \leq y \leq 2$: $x$ ranges from $y^2$ (from $y = \sqrt{x}$) to $4y$ (from $y = x/4$). For $2 < y \leq 4$: $x$ ranges from $4y$ to 16. Choice A has wrong upper limit in first integral. Choice C tries to use one integral but misses the complexity. Choice D has wrong lower limit in second integral.
3. The integral $\int_0^1 \int_{x^2}^{2x} \sin(y^3) \, dy \, dx$ is difficult to evaluate in the given order. After reversing the order of integration, what are the correct limits for the resulting integral?
   Answer: $\int_0^1 \int_{y/2}^{\sqrt{y}} \sin(y^3) \, dx \, dy + \int_1^2 \int_0^{y/2} \sin(y^3) \, dx \, dy$ The region is bounded by $y = x^2$ and $y = 2x$ for $0 \leq x \leq 1$. These curves intersect at $(0,0)$ and $(1,2)$. For $0 \leq y \leq 1$: $x$ ranges from $y/2$ to $\sqrt{y}$. For $1 < y \leq 2$: $x$ ranges from 0 to $y/2$. Choice A has wrong $x$ order. Choice B misses the split at $y = 1$. Choice C has incorrect limits in the second integral.
4. The integral $\int_0^{\pi} \int_0^{\sin x} y f(x,y) \, dy \, dx$ represents integration over a region bounded by the sine curve. A student claims this integral cannot have its order reversed because 'sine functions don't have nice inverse functions over the full domain.' Evaluate this claim.
   Answer: The student is incorrect; the integral can be reversed as $\int_0^1 \int_{\arcsin y}^{\pi - \arcsin y} y f(x,y) \, dx \, dy$ Over $[0,\pi]$, $\sin x$ ranges from 0 to 1 (maximum at $x = \pi/2$). For fixed $y \in [0,1]$, we need $\sin x \geq y$. This occurs when $\arcsin y \leq x \leq \pi - \arcsin y$ (using the symmetry of sine about $x = \pi/2$). The student's claim is wrong because we can use the arcsine function appropriately. Choice A accepts the incorrect claim. Choice C only uses one branch of the inverse. Choice D suggests unnecessary splitting.
5. Consider the region $R$ where $\iint_R f(x,y) \, dA = \int_{-1}^1 \int_{|x|}^1 f(x,y) \, dy \, dx$. When reversing to integrate with respect to $x$ first, which of the following correctly represents the region?
   Answer: $\int_0^1 \int_{-y}^y f(x,y) \, dx \, dy$ When you encounter a question about changing the order of integration, you're working with the fundamental skill of describing the same region in two different ways. The key is to carefully analyze the bounds and sketch the region to understand its geometry. The original integral $\int_{-1}^1 \int_{|x|}^1 f(x,y) \, dy \, dx$ describes a region where $x$ ranges from $-1$ to $1$, and for each $x$, $y$ ranges from $|x|$ to $1$. Since $|x| = x$ when $x \geq 0$ and $|x| = -x$ when $x < 0$, this creates a region bounded below by the V-shaped curve $y = |x|$ and above by the horizontal line $y = 1$. To reverse the order of integration, you need to express $x$ in terms of $y$. From the sketch, you can see that $y$ ranges from $0$ (the bottom of the V) to $1$ (the top line). For any fixed $y$ in this range, $x$ extends from the left branch of the V (where $y = -x$, so $x = -y$) to the right branch (where $y = x$, so $x = y$). This gives answer choice C: $\int_0^1 \int_{-y}^y f(x,y) \, dx \, dy$. Choice A incorrectly splits the region and extends $y$ beyond 1. Choice B uses constant bounds for $x$, ignoring the V-shape constraint. Choice D maintains the original order rather than reversing it. Study tip: Always sketch the region first when changing integration order. Identify the curves that bound your region, then carefully determine how the variables relate to each other in the new order.
6. Consider reversing the order of integration for $\int_0^2 \int_{x^2}^{4-x^2} f(x,y) \, dy \, dx$. The region is bounded by two parabolas. What is the key insight needed to determine where to split the reversed integral?
   Answer: Find where the parabolas $y = x^2$ and $y = 4-x^2$ intersect to determine the $y$-value that separates different $x$-ranges When reversing the order of integration in double integrals, you need to understand how the region's boundaries change when you switch from integrating $dy \, dx$ to $dx \, dy$. The key is identifying where different boundary functions apply as you move through your new outer variable's range. Looking at this integral, the region is bounded below by $y = x^2$ and above by $y = 4 - x^2$ for $x$ from 0 to 2. To reverse to $dx \, dy$, you need to express $x$ in terms of $y$, but here's the crucial insight: different parts of the region have different $x$-boundaries depending on the $y$-value you're at. The parabolas $y = x^2$ and $y = 4 - x^2$ intersect when $x^2 = 4 - x^2$, giving $x = \sqrt{2}$ and $y = 2$. This intersection point at $y = 2$ is critical because it determines where the region's structure changes. For $y$-values below 2, the region extends from the left branch of $y = x^2$ to the right branch. For $y$-values above 2, the region is bounded only by $y = 4 - x^2$. This forces you to split the integral at $y = 2$. Choice A focuses on $x$-limits rather than the splitting point. Choice B mentions vertices but misses that intersections, not vertices, determine where region structure changes. Choice D incorrectly suggests area calculations determine splitting necessity. Study tip: When reversing integration order, always find intersection points of boundary curves—they typically indicate where you need to split the integral because the boundary functions change roles.
7. The double integral $\int_0^{\pi/2} \int_0^{\cos x} f(x,y) \, dy \, dx$ represents integration over a region bounded by trigonometric curves. When reversing the order of integration, what is the most challenging aspect of determining the new limits?
   Answer: Finding the inverse function $x = \arccos(y)$ and determining that $y$ ranges from 0 to 1 with $x$ from $\arccos(y)$ to $\pi/2$ The region is bounded by $y = \cos x$ for $0 \leq x \leq \pi/2$ and $y \geq 0$. Since $\cos(0) = 1$ and $\cos(\pi/2) = 0$, $y$ ranges from 0 to 1. For fixed $y$, we need $x$ such that $\cos x = y$, giving $x = \arccos(y)$, and $x$ can range from $\arccos(y)$ to $\pi/2$. Choice B is wrong because no splitting is needed. Choice C uses wrong inverse function and range. Choice D is incorrect because the integral can be reversed.
8. Consider the double integral $\int_0^4 \int_{\sqrt{x}}^2 f(x,y) \, dy \, dx$. When the order of integration is reversed, which of the following represents the correct limits of integration?
   Answer: $\int_0^2 \int_0^{y^2} f(x,y) \, dx \, dy$ The region is bounded by $y = \sqrt{x}$, $y = 2$, $x = 0$, and $x = 4$. To reverse the order, we need $y$ from 0 to 2. For fixed $y$, $x$ ranges from 0 to $y^2$ (since $x = y^2$ from $y = \sqrt{x}$). Choice B incorrectly uses $x = y^2$ as the lower limit. Choice C doesn't reverse the order. Choice D uses incorrect $x$ limits.

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