Sketching Regions & Bounds - Multivariable Calculus

1. Consider the region $R$ bounded by the cylinder $x^2 + y^2 = 4$ and the planes $z = 0$, $z = 2$, and $x + y + z = 3$. Using cylindrical coordinates, which integral setup correctly computes the volume of $R$?
   Answer: $\int_0^{2\pi} \int_0^2 \int_{\max(0, 3-r\cos\theta-r\sin\theta)}^2 r \, dz \, dr \, d\theta$ The region is inside the cylinder $x^2 + y^2 = 4$ (so $0 \leq r \leq 2$) and bounded by three planes. In cylindrical coordinates: $z = 0$, $z = 2$, and $x + y + z = 3$ becomes $r\cos\theta + r\sin\theta + z = 3$, or $z = 3 - r\cos\theta - r\sin\theta$. The region exists where all constraints are satisfied simultaneously. For the $z$ bounds, we need $z \geq 0$, $z \leq 2$, and the plane constraint. The plane $x + y + z = 3$ can either provide an upper bound (when $3 - r\cos\theta - r\sin\theta < 2$) or a lower bound (when $3 - r\cos\theta - r\sin\theta > 0$). The lower bound for $z$ is $\max(0, 3 - r\cos\theta - r\sin\theta)$, and the upper bound is $2$. This is because when $3 - r\cos\theta - r\sin\theta > 0$, the plane is above $z = 0$, so we start from the plane. When $3 - r\cos\theta - r\sin\theta \leq 0$, we start from $z = 0$. Choice A uses min instead of max and has the wrong upper bound. Choice C ignores the plane constraint entirely. Choice D doesn't use the max function and could give negative lower bounds.
2. The region $R$ in the $xy$-plane is defined by $x^2 + y^2 \leq 9$, $x^2 + y^2 \geq 1$, and $x + y \geq 0$. Which of the following polar coordinate setups correctly represents $\iint_R f(x,y) \, dA$?
   Answer: $\int_{-\pi/4}^{3\pi/4} \int_1^3 f(r\cos\theta, r\sin\theta) \, r \, dr \, d\theta$ The region is an annulus ($1 \leq r \leq 3$) intersected with the half-plane $x + y \geq 0$. In polar coordinates, $x + y \geq 0$ becomes $r\cos\theta + r\sin\theta \geq 0$, which simplifies to $r(\cos\theta + \sin\theta) \geq 0$. Since $r > 0$, we need $\cos\theta + \sin\theta \geq 0$. This occurs when $\sin\theta \geq -\cos\theta$, or $\tan\theta \geq -1$. This inequality is satisfied when $\theta \in [-\pi/4 + 2\pi k, 3\pi/4 + 2\pi k]$ for integer $k$. Taking the principal range, we get $\theta \in [-\pi/4, 3\pi/4]$. The radial bounds are simply $1 \leq r \leq 3$. Choice B has the wrong angular range. Choice C includes regions where $x + y < 0$. Choice D has an incorrect angular range that misses part of the region.
3. Let $E$ be the solid region bounded by the parabolic cylinder $y=x^2$, the plane $y+z=4$, and the $xy$-plane ($z=0$). Which of the following iterated integrals represents the volume of $E$?
   Answer: $\int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} dz \, dy \, dx$ The solid $E$ is bounded below by the $xy$-plane, so $z=0$. It is bounded above by the plane $y+z=4$, so $z=4-y$. Thus, the bounds for $z$ are $0 \le z \le 4-y$. To determine the bounds for $x$ and $y$, we project the solid onto the $xy$-plane. The projection is the region bounded by $y=x^2$ and the intersection of $z=4-y$ with $z=0$, which is the line $y=4$. The region is described by $x^2 \le y \le 4$. The intersection of $y=x^2$ and $y=4$ occurs at $x=\pm 2$. Therefore, the bounds for the projection are $-2 \le x \le 2$ and $x^2 \le y \le 4$. Combining these in the order $dz \, dy \, dx$ gives the integral $\int_{-2}^{2} \int_{x^2}^{4} \int_{0}^{4-y} dz \, dy \, dx$.
4. The iterated integral $I = \int_0^1 \int_{\arcsin y}^{\pi - \arcsin y} f(x,y) \, dx \, dy$ is taken over a region $R$. Which of the following accurately describes the region $R$?
   Answer: The region bounded by the $x$-axis and the curve $y=\sin(x)$ for $0 \le x \le \pi$. The bounds of integration are $0 \le y \le 1$ and $\arcsin y \le x \le \pi - \arcsin y$. Let's analyze the bounds for $x$. The left boundary is $x = \arcsin y$, which is equivalent to $y = \sin x$ for $x \in [0, \pi/2]$. The right boundary is $x = \pi - \arcsin y$. If we take the sine of both sides, $\sin(x) = \sin(\pi - \arcsin y) = \sin(\arcsin y) = y$. This corresponds to the part of the sine curve where $x \in [\pi/2, \pi]$. Thus, for a fixed $y$ between 0 and 1, $x$ ranges from the ascending part of the sine curve to the descending part. The overall region $R$ is bounded above by $y=\sin x$ and below by the $x$-axis ($y=0$) over the interval $x \in [0, \pi]$.
5. Consider the iterated integral $I = \int_0^1 \int_{\sqrt{y}}^{2-y} f(x,y) \, dx \, dy$. Which of the following expressions is equivalent to $I$ when the order of integration is reversed?
   Answer: $\int_0^1 \int_0^{x^2} f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx$ The region of integration is defined by $0 \le y \le 1$ and $\sqrt{y} \le x \le 2-y$. The left boundary is $x=\sqrt{y}$ (or $y=x^2$ for $x \ge 0$) and the right boundary is $x=2-y$ (or $y=2-x$). The intersection of these curves occurs when $x^2 = 2-x$, which gives $x^2+x-2=0$, or $(x+2)(x-1)=0$. Since $x=\sqrt{y} \ge 0$, the intersection is at $x=1$, which corresponds to $y=1$. To reverse the order of integration, we must express the bounds of $y$ as functions of $x$. The region must be split at $x=1$. For $0 \le x \le 1$, the region is bounded below by $y=0$ and above by the parabola $y=x^2$. For $1 \le x \le 2$, the region is bounded below by $y=0$ and above by the line $y=2-x$. This leads to the sum of two integrals: $\int_0^1 \int_0^{x^2} f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx$.
6. Let $R$ be the region in the $xy$-plane that is inside the circle $x^2+y^2=2x$ but outside the circle $x^2+y^2=1$. Which of the following integrals represents the area of $R$?
   Answer: $\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} r \, dr \, d\theta$ When you encounter a region bounded by circles in polar coordinates, your first step is converting the Cartesian equations to polar form and identifying where the boundaries intersect. The circle $x^2 + y^2 = 2x$ can be rewritten by completing the square: $(x-1)^2 + y^2 = 1$. This is a circle centered at $(1,0)$ with radius 1. In polar coordinates, substituting $x = r\cos\theta$ gives us $r^2 = 2r\cos\theta$, so $r = 2\cos\theta$. The circle $x^2 + y^2 = 1$ becomes $r = 1$ in polar coordinates. To find the intersection points, set $1 = 2\cos\theta$, giving $\cos\theta = \frac{1}{2}$, so $\theta = \pm\frac{\pi}{3}$. This means the region extends from $\theta = -\frac{\pi}{3}$ to $\theta = \frac{\pi}{3}$. For any fixed angle $\theta$ in this range, $r$ varies from the inner boundary ($r = 1$) to the outer boundary ($r = 2\cos\theta$). The area element in polar coordinates is $r \, dr \, d\theta$, giving us $\int_{-\pi/3}^{\pi/3} \int_{1}^{2\cos\theta} r \, dr \, d\theta$. Option A has the integration limits reversed for $r$. Option B uses incorrect $\theta$ limits—at $\theta = \pm\frac{\pi}{2}$, the outer circle has $r = 0$, which doesn't make sense for our region. Option D uses the wrong area element (missing the $r$). Study tip: Always sketch the region first, find intersection points to determine your limits, and remember that polar area elements include the factor $r$.
7. A solid region $E$ lies above the cone $z = \sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=z$. Which of the following integrals in spherical coordinates represents the volume of $E$?
   Answer: $\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$ First, convert the boundary equations to spherical coordinates. The cone $z = \sqrt{x^2+y^2}$ becomes $\rho\cos\phi = \sqrt{(\rho\sin\phi)^2} = \rho\sin\phi$, which simplifies to $\tan\phi=1$, so $\phi=\pi/4$. The region is 'above' the cone, meaning angles are between the $z$-axis and the cone, so $0 \le \phi \le \pi/4$. The sphere $x^2+y^2+z^2=z$ becomes $\rho^2 = \rho\cos\phi$, which simplifies to $\rho=\cos\phi$. This defines the outer boundary for $\rho$. The region is symmetric about the $z$-axis, so $0 \le \theta \le 2\pi$. The volume element in spherical coordinates is $dV = \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$. Therefore, the volume is given by the integral $\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\cos\phi} \rho^2\sin\phi \, d\rho \, d\phi \, d\theta$.
8. Let $R$ be the trapezoidal region in the $xy$-plane with vertices $(1,0), (2,0), (0,2), (0,1)$. Using the transformation $u=x+y$ and $v=y-x$, the integral $\iint_R (x+y) \, dA$ is transformed into which of the following integrals over a region $S$ in the $uv$-plane?
   Answer: $\int_1^2 \int_{-u}^{u} \frac{1}{2} u \, dv \, du$ First, we transform the boundaries of $R$. The vertices define the lines $y=0$, $x=0$, $x+y=1$, and $x+y=2$. In terms of $u$ and $v$: $x+y=u$ means two boundaries are $u=1$ and $u=2$. The line $y=0$ means $u=x, v=-x$, so $v=-u$. The line $x=0$ means $u=y, v=y$, so $v=u$. The region $S$ in the $uv$-plane is bounded by $u=1, u=2, v=-u, v=u$. Next, we find the Jacobian of the transformation from $(u,v)$ to $(x,y)$. We need $x$ and $y$ in terms of $u$ and $v$. Adding the transformations gives $u+v=2y \implies y=(u+v)/2$. Subtracting gives $u-v=2x \implies x=(u-v)/2$. The Jacobian determinant is $J = \det \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix} = \det \begin{pmatrix} 1/2 & -1/2 \\ 1/2 & 1/2 \end{pmatrix} = 1/4 - (-1/4) = 1/2$. The integrand $(x+y)$ becomes $u$. So the integral is $\iint_S u |J| \, dA_{uv} = \int_1^2 \int_{-u}^{u} u \cdot \frac{1}{2} \, dv \, du$.
9. Let $R$ be the region in the $xy$-plane that lies inside the cardioid $r = 1 + \cos\theta$ and outside the circle $r=1$. Which iterated integral in polar coordinates gives the volume of the solid under the surface $z=y$ and above the region $R$?
   Answer: $\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r^2 \sin\theta \, dr \, d\theta$ The region of integration $R$ is defined by the bounds of $r$ and $\theta$. The curves intersect when $1 = 1 + \cos\theta$, which means $\cos\theta = 0$. This occurs at $\theta = -\pi/2$ and $\theta = \pi/2$. So, the bounds for $\theta$ are from $-\pi/2$ to $\pi/2$. For any such $\theta$, $r$ ranges from the inner circle $r=1$ to the outer cardioid $r=1+\cos\theta$. The volume is given by the double integral of the height function $z=y$ over $R$. In polar coordinates, $y=r\sin\theta$, and the area element is $dA = r \, dr \, d\theta$. The integrand is thus $(r\sin\theta) \cdot r = r^2\sin\theta$. Combining these gives the integral $\int_{-\pi/2}^{\pi/2} \int_1^{1+\cos\theta} r^2 \sin\theta \, dr \, d\theta$.
10. Let $E$ be the solid tetrahedron with vertices at $(0,0,0), (2,0,0), (0,4,0),$ and $(0,0,6)$. Which of the following iterated integrals gives the volume of $E$?
    Answer: $\int_0^2 \int_0^{4-2x} \int_0^{6-3x-3y/2} dz \, dy \, dx$ The four vertices lie on the coordinate planes except for the three axial intercepts $(2,0,0), (0,4,0), (0,0,6)$. The plane passing through these three points has the intercept form equation $\frac{x}{2} + \frac{y}{4} + \frac{z}{6} = 1$. Solving for $z$ gives the upper bound for the solid: $z = 6 - 3x - \frac{3}{2}y$. The lower bound is the $xy$-plane, $z=0$. The projection of the tetrahedron onto the $xy$-plane is a triangle with vertices $(0,0), (2,0), (0,4)$. The line connecting $(2,0)$ and $(0,4)$ is given by $\frac{x}{2} + \frac{y}{4} = 1$, or $y=4-2x$. To integrate with respect to $y$ then $x$, the bounds are $0 \le y \le 4-2x$ for $0 \le x \le 2$. Combining these gives the integral $\int_0^2 \int_0^{4-2x} \int_0^{6-3x-3y/2} dz \, dy \, dx$.
11. Let $E$ be the solid region bounded by the paraboloid $z=x^2+y^2$ and the plane $z=2x+3$. Which of the following describes the projection of $E$ onto the $xy$-plane?
    Answer: The disk defined by $(x-1)^2+y^2 \le 4$. The projection of the solid region $E$ onto the $xy$-plane is determined by the intersection of its bounding surfaces. We set the $z$-values equal to find the curve of intersection: $x^2+y^2 = 2x+3$. To identify this curve, we can complete the square for the $x$ terms: $x^2 - 2x + y^2 = 3 \implies (x^2 - 2x + 1) + y^2 = 3 + 1 \implies (x-1)^2 + y^2 = 4$. This is the equation of a circle in the $xy$-plane centered at $(1,0)$ with a radius of $2$. The projection is the region enclosed by this circle, which is the disk defined by the inequality $(x-1)^2+y^2 \le 4$.
12. The region $R$ is bounded by the curves $xy = 1$, $xy = 4$, $y = x$, and $y = 4x$. Using the transformation $u = xy$ and $v = y/x$, what are the bounds for the transformed region in the $uv$-plane?
    Answer: $1 \leq u \leq 4, \quad 1 \leq v \leq 4$ Under the transformation $u = xy$ and $v = y/x$, let's see how each boundary curve transforms: The curves $xy = 1$ and $xy = 4$ become $u = 1$ and $u = 4$ respectively. The curve $y = x$ gives $v = y/x = x/x = 1$. The curve $y = 4x$ gives $v = y/x = 4x/x = 4$. Therefore, the transformed region is $1 \leq u \leq 4$ and $1 \leq v \leq 4$. Choice B incorrectly computes the $v$ bounds by taking reciprocals. Choice C incorrectly limits the upper $v$ bound. Choice D swaps the $u$ and $v$ ranges incorrectly.
13. Let $R$ be the region in the first quadrant defined by the inequalities $1 \le x^2+y^2 \le 4$ and $y \le x \le \sqrt{3}y$. Which of the following integrals in polar coordinates represents the area of $R$?
    Answer: $\int_{\pi/6}^{\pi/4} \int_1^2 r \, dr \, d\theta$ When converting a region from Cartesian to polar coordinates, you need to carefully translate each boundary condition and remember that area integrals in polar coordinates always include the Jacobian factor $r$. Let's analyze each boundary of region $R$. The condition $1 \le x^2+y^2 \le 4$ becomes $1 \le r^2 \le 4$, so $1 \le r \le 2$. The linear boundaries $y \le x$ and $x \le \sqrt{3}y$ require converting to polar form using $x = r\cos\theta$ and $y = r\sin\theta$. For $y \le x$: $r\sin\theta \le r\cos\theta$, which simplifies to $\tan\theta \le 1$, giving us $\theta \le \pi/4$. For $x \le \sqrt{3}y$: $r\cos\theta \le \sqrt{3}r\sin\theta$, which simplifies to $\cot\theta \le \sqrt{3}$, so $\tan\theta \ge 1/\sqrt{3}$, giving us $\theta \ge \pi/6$. Therefore, $\pi/6 \le \theta \le \pi/4$ and $1 \le r \le 2$. The area integral is $\int_{\pi/6}^{\pi/4} \int_1^2 r \, dr \, d\theta$, which is choice C. Choice A has the wrong $r$ limits (4 instead of 2). Choice B uses incorrect $\theta$ limits ($\pi/4$ to $\pi/3$ instead of $\pi/6$ to $\pi/4$). Choice D is missing the crucial Jacobian factor $r$ in the integrand. Key strategy: Always sketch the region first, then systematically convert each boundary. Remember that area in polar coordinates requires the factor $r$, and double-check your angle conversions using basic trigonometric values.
14. A solid region $E$ is bounded below by the cone $z=\sqrt{x^2+y^2}$ and above by the paraboloid $z=2-x^2-y^2$. Which of the following integrals in cylindrical coordinates represents the volume of $E$?
    Answer: $\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r^2} r \, dz \, dr \, d\theta$ In cylindrical coordinates, the cone $z=\sqrt{x^2+y^2}$ becomes $z=r$. The paraboloid $z=2-x^2-y^2$ becomes $z=2-r^2$. The solid is bounded below by $z=r$ and above by $z=2-r^2$. To find the bounds for $r$, we find the intersection of these surfaces: $r = 2-r^2 \implies r^2+r-2=0 \implies (r+2)(r-1)=0$. Since radius $r$ must be non-negative, $r=1$. This means the projection of the solid onto the $xy$-plane is a disk of radius 1. So, $0 \le r \le 1$. The solid is symmetric about the $z$-axis, so $0 \le \theta \le 2\pi$. The volume element is $dV = r \, dz \, dr \, d\theta$. The correct setup is $\int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r^2} r \, dz \, dr \, d\theta$.
15. Consider the region in the first octant bounded by the coordinate planes and the plane $2x + 3y + z = 6$. When integrating in the order $dz \, dy \, dx$, what are the correct bounds?
    Answer: $\int_0^3 \int_0^{(6-2x)/3} \int_0^{6-2x-3y} dz \, dy \, dx$ The region is in the first octant, so $x \geq 0$, $y \geq 0$, and $z \geq 0$. The plane $2x + 3y + z = 6$ intersects the coordinate axes at $(3,0,0)$, $(0,2,0)$, and $(0,0,6)$. For the outermost integral ($dx$): $x$ ranges from $0$ to $3$ (where the plane intersects the $x$-axis). For the middle integral ($dy$): Given a fixed $x$, $y$ ranges from $0$ to where the plane intersects the $xz$-plane at that $x$ value. Setting $z = 0$ in the plane equation: $2x + 3y = 6$, so $y = (6-2x)/3$. For the innermost integral ($dz$): Given fixed $x$ and $y$, $z$ ranges from $0$ to $z = 6 - 2x - 3y$ (from the plane equation). Choice B has incorrect upper bound for $x$ and wrong middle bounds. Choice C uses $6-2x$ instead of $(6-2x)/3$ for the $y$ bound. Choice D uses $2$ instead of $3$ as the upper $x$ bound.

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