Sketching Regions & Surfaces - Calculus 3

1. The region of integration for a double integral is described by $1 ≤ x ≤ 3$ and $\ln x ≤ y ≤ e^x$. When sketching this region, which geometric feature is most important for determining the integration order?
   Answer: The exponential growth of $y = e^x$ compared to the logarithmic growth of $y = \ln x$ creates a region where switching to $dy \, dx$ order would require splitting into multiple integrals To analyze this region, we note that for $x ∈ [1, 3]$, we have $y$ ranging from $\ln x$ to $e^x$. At $x = 1$: $\ln 1 = 0$ and $e^1 = e ≈ 2.718$. At $x = 3$: $\ln 3 ≈ 1.099$ and $e^3 ≈ 20.086$. So $y$ ranges from 0 to approximately 20. If we switch to $dy \, dx$ order, we need to express $x$ as a function of $y$. For $y = \ln x$, we get $x = e^y$. For $y = e^x$, we get $x = \ln y$. However, these inverse relationships are valid over different $y$-intervals, requiring us to split the region. For small $y$-values (roughly $0 ≤ y ≤ 1$), $x$ ranges from $e^y$ to $\ln y$ (but $\ln y$ undefined for $y ≤ 0$). The complexity of describing $x$-limits as functions of $y$ makes the $dx \, dy$ order much more difficult. Choice A is wrong because the curves don't intersect in $[1,3]$. Choice C incorrectly suggests horizontal strips are more complex. Choice D wrongly claims equal complexity for both orders.
2. Consider the region $R$ bounded by $r = 1 + \cos \theta$ in polar coordinates, where $0 ≤ \theta ≤ 2\pi$. When converting this to a double integral in rectangular coordinates, which characteristic of this region presents the greatest challenge?
   Answer: The polar equation $r = 1 + \cos \theta$ produces a region where each $x$-value corresponds to up to four different $y$-values, requiring the rectangular integration to be split into multiple subregions The cardioid $r = 1 + \cos \theta$ has the characteristic that for most $x$-values in its domain, there are multiple corresponding $y$-values. To see this, consider a vertical line at some $x = c$ where $-2 < c < 2$. This line will generally intersect the cardioid at four points (except near the cusp and the rightmost point). Converting to rectangular coordinates requires solving $x = r\cos \theta = (1 + \cos \theta)\cos \theta$ and $y = r\sin \theta = (1 + \cos \theta)\sin \theta$ for $\theta$ in terms of $x$ and $y$, which yields multiple solutions for most $x$-values. This necessitates breaking the region into multiple pieces for integration in rectangular coordinates. Choice A correctly identifies the cardioid and cusp but doesn't emphasize the main integration challenge. Choice B incorrectly describes the symmetry (cardioids are symmetric about the $x$-axis). Choice D correctly identifies key points but misses the multiple-valued nature that creates the primary difficulty.
3. Consider the region in the $xy$-plane bounded by $y = \sin x$, $y = 0$, $x = 0$, and $x = 2\pi$. When setting up a double integral $\iint_R f(x,y) \, dA$ over this region, which feature most significantly complicates the choice of integration order?
   Answer: If we integrate in the order $dy \, dx$, the region can be described simply, but integrating in the order $dx \, dy$ requires solving $x = \arcsin y$ which has multiple solutions in $[0, 2\pi]$ When setting up double integrals, you must carefully consider how the region's boundary equations affect your integration limits, especially when switching the order of integration. Let's analyze this region bounded by $y = \sin x$, $y = 0$, $x = 0$, and $x = 2\pi$. For the order $dy \, dx$, you can describe the region straightforwardly: $0 \leq x \leq 2\pi$ and $0 \leq y \leq \sin x$ (taking the absolute value or splitting at $x = \pi$ to handle the negative portion). However, if you want to integrate in the order $dx \, dy$, you need to express $x$ as a function of $y$. This means solving $y = \sin x$ for $x$, giving $x = \arcsin y$. Here's the critical issue: within the interval $[0, 2\pi]$, the equation $\sin x = y$ has multiple solutions for most values of $y$ in $[0,1]$. Specifically, if $y = \sin x$, then $x$ could equal $\arcsin y$ or $\pi - \arcsin y$ (and potentially more values), making it impossible to describe the region with simple inequalities. Choice A incorrectly focuses on the negative values as the primary complication. Choice B wrongly suggests the region splits into separate components. Choice C incorrectly claims there are three lobes when there are actually two symmetric parts above and below the x-axis. The answer is D because the multiple solutions to $x = \arcsin y$ make the $dx \, dy$ order significantly more complex to set up properly. Study tip: Always check whether boundary functions are one-to-one when considering integration order—many-to-one relationships create the most complications.
4. A region $D$ in the first octant is bounded by the coordinate planes, the plane $x + y + z = 6$, and the cylinder $x^2 + y^2 = 9$. When sketching this region for a triple integral, what is the most restrictive constraint on the domain?
   Answer: The cylinder $x^2 + y^2 = 9$ limits the projection onto the $xy$-plane to a quarter-circle of radius 3, and this constraint is never overridden by the plane $x + y + z = 6$ To analyze this region, we examine the constraints systematically. In the first octant, we have $x ≥ 0$, $y ≥ 0$, $z ≥ 0$. The cylinder $x^2 + y^2 = 9$ (radius 3) restricts the $xy$-projection to a quarter-circle. The plane $x + y + z = 6$ provides an upper bound: $z ≤ 6 - x - y$. For this to be meaningful, we need $6 - x - y ≥ 0$, or $x + y ≤ 6$. Now we check: does the constraint $x + y ≤ 6$ further restrict the quarter-circle $x^2 + y^2 ≤ 9$? The maximum value of $x + y$ on the quarter-circle occurs at $(3/\sqrt{2}, 3/\sqrt{2})$ where $x + y = 6/\sqrt{2} = 3\sqrt{2} ≈ 4.24 < 6$. Therefore, the constraint $x + y ≤ 6$ is never violated within the cylindrical region, making the cylinder the most restrictive constraint. Choice B incorrectly suggests the plane creates a smaller projection. Choice C correctly notes the line $x + y = 6$ but wrongly implies it restricts the region. Choice D underestimates the importance of the cylinder constraint.
5. The surface $z = \sqrt{x^2 + y^2}$ is intersected by the planes $z = 2$ and $z = 5$. When sketching the resulting solid region for integration purposes, which statement best describes the geometric relationship between these boundaries?
   Answer: The region is the portion of a cone $z = \sqrt{x^2 + y^2}$ between two horizontal planes, creating a frustum with circular cross-sections of radius $r = z$ at each height $z$ The surface $z = \sqrt{x^2 + y^2}$ represents a cone opening upward with vertex at the origin. At any height $z = h$, the cross-section is a circle with radius $r = h$ (since $\sqrt{x^2 + y^2} = h$ gives $x^2 + y^2 = h^2$). When this cone is intersected by the planes $z = 2$ and $z = 5$, we get the portion of the cone between these heights. At $z = 2$, the radius is 2; at $z = 5$, the radius is 5. This creates a frustum (truncated cone) where the radius equals the height at each level. Choice A incorrectly describes an annular cylinder (which would have constant inner and outer radii). Choice B correctly identifies a truncated cone but incorrectly describes it as having the form of a different surface. Choice D incorrectly suggests two separate regions rather than a single connected frustum.
6. Consider the region $R$ bounded by the curves $y = x^2 + 1$, $y = 5 - x^2$, and the vertical lines $x = -1$ and $x = 1$. Which statement best describes the shape and characteristics of this region for setting up a double integral?
   Answer: The region is bounded above by $y = 5 - x^2$ and below by $y = x^2 + 1$ for $-1 ≤ x ≤ 1$, forming a lens-shaped region with maximum height at $x = 0$ To determine the region, we need to identify which curve is above the other. At any point in $[-1, 1]$, we compare $y = x^2 + 1$ and $y = 5 - x^2$. Since $5 - x^2 ≥ x^2 + 1$ when $2x^2 ≤ 4$ or $x^2 ≤ 2$, and this is satisfied for all $x ∈ [-1, 1]$, we have $5 - x^2 > x^2 + 1$ throughout this interval. The maximum height occurs at $x = 0$ where the distance between curves is $(5 - 0) - (0 + 1) = 4$. The minimum height occurs at $x = ±1$ where the distance is $(5 - 1) - (1 + 1) = 2$. This creates a lens-shaped region. Choice B incorrectly reverses which curve is on top. Choice C is wrong because the curves don't need to intersect for the region to be well-defined. Choice D incorrectly states the height is minimum at $x = 0$.
7. Consider the surface $z = xy$ and the region $R$ in the $xy$-plane bounded by $y = x^2$ and $y = 4$. When sketching this surface over region $R$ to evaluate $\iint_R f(x,y) \, dA$, which description best characterizes the behavior of the surface?
   Answer: The surface has both positive and negative values over region $R$, with the surface below the $xy$-plane when $x < 0$ and above when $x > 0$, and zero along the line $x = 0$ First, we identify region $R$: it's bounded by $y = x^2$ (below) and $y = 4$ (above). The parabola $y = x^2$ intersects $y = 4$ when $x^2 = 4$, so $x = ±2$. Thus $R = \{(x,y) : -2 ≤ x ≤ 2, x^2 ≤ y ≤ 4\}$. The surface $z = xy$ takes both positive and negative values: when $x > 0$, since $y ≥ x^2 ≥ 0$, we have $z = xy ≥ 0$; when $x < 0$, since $y ≥ x^2 > 0$, we have $z = xy < 0$; when $x = 0$, we have $z = 0$. Choice A incorrectly focuses on the saddle point at origin (which isn't in region $R$) and mentions wrong maximum locations. Choice B incorrectly claims the surface is always positive. Choice D correctly notes symmetry but incorrectly describes where extrema occur.
8. A solid region $S$ is defined by $x^2 + y^2 ≤ 4$, $0 ≤ z ≤ 6 - x - y$. When sketching this region to set up a triple integral, what is the most significant geometric constraint that determines the actual shape of $S$?
   Answer: The constraint $6 - x - y ≥ 0$ limits the base region to the portion of the disk where $x + y ≤ 6$, but since the disk has radius 2, this constraint is never binding for the integration To understand this region, we need to determine where $6 - x - y ≥ 0$, i.e., where $x + y ≤ 6$. The disk $x^2 + y^2 ≤ 4$ has radius 2, so points in this disk satisfy $-2 ≤ x ≤ 2$ and $-2 ≤ y ≤ 2$. The maximum value of $x + y$ on this disk occurs at $(2, 2)$ giving $x + y = 4$, and the minimum occurs at $(-2, -2)$ giving $x + y = -4$. Since $4 < 6$, the constraint $x + y ≤ 6$ is satisfied throughout the entire disk, making it non-binding. The region is simply a cylinder with base $x^2 + y^2 ≤ 4$ and height varying from $z = 0$ to $z = 6 - x - y$. Choice B incorrectly suggests the constraint truncates the region. Choice C wrongly implies the line $x + y = 6$ intersects the disk. Choice D incorrectly states that $z = 6 - x - y$ becomes negative within the disk.
9. The solid region $W$ is bounded by the paraboloid $z = x^2 + y^2$, the cylinder $x^2 + y^2 = 4$, and the plane $z = 8$. When sketching this region for a triple integral in cylindrical coordinates, which statement best describes the geometric constraints?
   Answer: The region is bounded below by $z = r^2$ and above by $z = 8$, but only for $0 ≤ r ≤ 2$, with the constraint that $r^2 ≤ 8$ being automatically satisfied within the cylindrical boundary When setting up triple integrals in cylindrical coordinates for regions bounded by multiple surfaces, you need to carefully analyze how each boundary constrains the integration limits. Let's examine each surface in cylindrical coordinates where $x = r\cos\theta$, $y = r\sin\theta$. The paraboloid $z = x^2 + y^2$ becomes $z = r^2$, the cylinder $x^2 + y^2 = 4$ becomes $r = 2$, and the plane remains $z = 8$. The key insight is determining which surfaces actually bound your region. The cylinder $r = 2$ limits the radial extent, so $0 \leq r \leq 2$. Within this cylindrical boundary, the paraboloid $z = r^2$ serves as the lower surface, while the plane $z = 8$ serves as the upper surface. Since the maximum value of $r^2$ when $r = 2$ is only 4, and $4 < 8$, the condition $r^2 \leq 8$ is automatically satisfied for all points within the cylinder. This makes answer D correct. A is wrong because it suggests the region extends from the $xy$-plane, but the paraboloid $z = r^2$ is the actual lower boundary, not $z = 0$. B incorrectly focuses on where the paraboloid intersects $z = 8$ at $r = 2\sqrt{2} \approx 2.83$, but this intersection point lies outside our cylindrical constraint $r \leq 2$. C makes the same error by emphasizing the condition $r < 2\sqrt{2}$, which is irrelevant since the cylinder already restricts $r \leq 2$. Strategy tip: Always check which constraints are actually active—sometimes one boundary condition automatically satisfies another, making certain conditions redundant in your final setup.
10. A region $S$ is defined by $x^2 + y^2 + z^2 ≤ 16$, $z ≥ 0$, and $x + y + z ≥ 2$. When sketching this region, the intersection of the sphere and plane creates a boundary curve. What is the most accurate description of how this boundary curve affects the shape of region $S$?
    Answer: The intersection creates an elliptical boundary curve, and the plane removes a significant cap-shaped portion from the hemisphere, with the removed portion including the lowest point of the hemisphere When analyzing the intersection of multiple 3D constraints, you need to understand how each boundary affects the overall region and determine which constraint is most restrictive in different areas. The region S is bounded by three conditions: a sphere of radius 4, the upper half-space (z ≥ 0), and the half-space above the plane x + y + z = 2. To understand the shape, examine where the plane intersects the hemisphere. The sphere has equation x² + y² + z² = 16, and substituting into the plane equation x + y + z = 2 gives the intersection curve. The plane x + y + z = 2 passes through points like (2,0,0), (0,2,0), and (0,0,2) - all well inside the sphere since these points have distance √4 = 2 from the origin, which is less than the radius 4. This means the plane cuts significantly into the lower portion of the hemisphere. Option A is incorrect because the intersection is not a "small" portion - the plane removes a substantial cap-shaped region. Option C is wrong because the plane x + y + z = 2 does not pass through the sphere's center at (0,0,0), so it cannot divide the hemisphere into equal portions. Option D is false because the plane is not tangent to the sphere; it intersects the sphere in a full curve, not at a single point. Option B correctly describes that the intersection creates a circular cross-section (which may appear elliptical from certain viewing angles), and the plane removes a significant cap-shaped portion including the hemisphere's lowest point (0,0,0). Study tip: When visualizing intersections of spheres and planes, always check whether the plane passes through, near, or far from the center to gauge how much volume is affected.

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