Triple Integrals: Cartesian - Multivariable Calculus

1. Let $E$ be the solid region bounded by the cylinder $y^2+z^2=1$ and the planes $x=0$ and $x=y$. Evaluate the integral $\iiint_E z \, dV$.
   Answer: $0$ The region of integration is over the disk $D$ in the $yz$-plane, which is $y^2+z^2 \le 1$. The bounds for $x$ are between the planes $x=0$ and $x=y$. This means for $y>0$, $0 \le x \le y$, and for $y<0$, $y \le x \le 0$. We can set up the integral as $\iiint_E z \, dV = \iint_D \left(\int_{x_{min}}^{x_{max}} z \, dx\right) dA_{yz}$. The inner integral is $\int z \, dx = zx$. Evaluating this gives $zy$ for $y>0$ and $-zy$ for $y<0$. So, the integral is $\iint_D |y|z \, dA_{yz}$. This can be split into two parts for $y>0$ and $y<0$. Let's integrate with respect to $y$ first: $\int_{-1}^1 \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} |y|z \, dy \, dz$. For any fixed $z$, the inner integral $\int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} |y|z \, dy = z \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} |y| \, dy$. Since $|y|$ is an even function, the integral $\int_{-a}^a |y| dy$ is positive. However, we then integrate the result, which is of the form $C \cdot z$ where $C>0$, with respect to $z$ from $-1$ to $1$. The function $Cz$ is an odd function of $z$, and the interval $[-1,1]$ is symmetric about $z=0$. Therefore, the outer integral $\int_{-1}^1 Cz \, dz = 0$.
2. Calculate the volume of the solid bounded by the planes $x=0$, $z=0$, $y=2x$, $y=4$, and $x+z=1$.
   Answer: $\frac{4}{3}$ When you encounter a problem asking for the volume of a solid bounded by multiple planes, you're dealing with a triple integral setup. The key is carefully determining the region of integration by analyzing how these five planes intersect to form a bounded solid. Start by sketching the region. The planes $x=0$, $z=0$, $y=2x$, $y=4$, and $x+z=1$ create a solid in the first octant. From the constraints, $x$ ranges from 0 to where $y=2x$ meets $y=4$, so $0 \leq x \leq 2$. For each $x$, $y$ goes from $2x$ to $4$, and $z$ ranges from 0 to $1-x$ (from the plane $x+z=1$). Set up the triple integral: $V = \int_0^2 \int_{2x}^4 \int_0^{1-x} dz \, dy \, dx$ Working from inside out: - $\int_0^{1-x} dz = 1-x$ - $\int_{2x}^4 (1-x) dy = (1-x)(4-2x) = (1-x) \cdot 2(2-x) = 2(1-x)(2-x)$ - $\int_0^2 2(1-x)(2-x) dx = 2\int_0^2 (2-3x+x^2) dx = 2[2x - \frac{3x^2}{2} + \frac{x^3}{3}]_0^2 = 2(4-6+\frac{8}{3}) = \frac{4}{3}$ Answer choice A ($\frac{8}{3}$) likely comes from forgetting the factor of 2 in the $y$-integration. Answer B (2) and D (4) probably result from integration errors or incorrect bounds setup. Strategy tip: Always sketch the region first and double-check your bounds by testing corner points. Triple integrals require careful attention to the order of integration and corresponding limits.
3. Let $I = \int_0^2 \int_0^{4-x^2} \frac{xe^{2y}}{4-y} dy \, dx$. What is the value of the integral $I$?
   Answer: $\frac{e^8 - 1}{4}$ When you encounter a double integral where the integrand has a potential singularity within the region of integration, you need to carefully examine whether the integral converges or diverges. Let's analyze the region of integration first. For $x \in [0,2]$, we have $y \in [0, 4-x^2]$. The integrand is $\frac{xe^{2y}}{4-y}$, which has a potential singularity when $4-y = 0$, or $y = 4$. We need to check if $y = 4$ lies within our region. When $x = 0$, $y$ ranges from $0$ to $4$, so the singularity occurs at the boundary point $(0,4)$. To evaluate this integral, we integrate with respect to $y$ first. Let $u = 4-y$, so $du = -dy$ and $e^{2y} = e^{2(4-u)} = e^8 \cdot e^{-2u}$. When $y = 0$, $u = 4$; when $y = 4-x^2$, $u = x^2$. The integral becomes: $I = \int_0^2 x e^8 \int_4^{x^2} \frac{e^{-2u}}{u} (-du) dx = \int_0^2 x e^8 \int_{x^2}^4 \frac{e^{-2u}}{u} du \, dx$ This integral converges because the singularity at $u = 0$ is outside our integration region, and $\int_{x^2}^4 \frac{e^{-2u}}{u} du$ can be evaluated to yield $I = \frac{e^8 - 1}{4}$. Choice (A) is wrong because the integral converges despite the apparent singularity. Choices (B) and (C) result from computational errors in handling the exponential terms or the change of variables incorrectly. Study tip: Always check if apparent singularities actually lie within your integration region, and consider substitutions to simplify complex integrands.
4. What is the average value of the function $f(x,y,z) = 6xz$ over the solid tetrahedron with vertices at $(0,0,0)$, $(1,0,0)$, $(0,2,0)$, and $(0,0,3)$?
   Answer: $\frac{9}{10}$ The plane passing through $(1,0,0)$, $(0,2,0)$, and $(0,0,3)$ is given by $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$, or $6x+3y+2z=6$. The volume of the tetrahedron is $V = \frac{1}{6} \cdot 1 \cdot 2 \cdot 3 = 1$. The average value is $\frac{1}{V} \iiint_E f(x,y,z) dV$. We compute $\iiint_E 6xz \, dV = \int_0^1 \int_0^{2(1-x)} \int_0^{3(1-x-y/2)} 6xz \, dz \, dy \, dx$. The inner integral with respect to $z$ yields $3x(3(1-x-y/2))^2 = 27x(1-x-y/2)^2$. The next integral with respect to $y$ is $\int_0^{2(1-x)} 27x(1-x-y/2)^2 \, dy = 18x(1-x)^3$. The outer integral is $\int_0^1 18x(1-x)^3 \, dx = 18 \cdot \frac{1!3!}{5!} = \frac{18}{20} = \frac{9}{10}$. Since the volume is 1, the average value is $\frac{9}{10}$.
5. The region $W$ is defined by $1 \leq x^2 + y^2 + z^2 \leq 4$, $z \geq 0$, and $x + y + z \leq 3$. If we want to compute $\iiint_W z^2 \, dV$ using Cartesian coordinates, which of the following describes the most significant computational challenge?
   Answer: The spherical bounds $x^2 + y^2 + z^2 \leq 4$ and $x^2 + y^2 + z^2 \geq 1$ create complex radical expressions in all six orders of integration The spherical constraints $1 \leq x^2 + y^2 + z^2 \leq 4$ create bounds like $\sqrt{1 - x^2 - y^2} \leq z \leq \sqrt{4 - x^2 - y^2}$ (when integrating $dz$ last), or similar radical expressions for other variables when integrating them last. These radical expressions make the antiderivatives complex regardless of the order chosen. The constraint $x + y + z \leq 3$ adds another layer but doesn't make setup impossible - it just adds another bound to consider. Choice B is incorrect because all orders are possible to set up, just computationally intensive. Choice C is wrong because $z^2$ doesn't require integration by parts - the difficulty comes from the bounds, not the integrand. Choice D is incorrect because $W$ is connected (it's a portion of a spherical shell).
6. A solid wedge is cut from the cylinder $x^2+y^2=1$ by the planes $z=0$ and $z=y$ in the region where $y \ge 0$. What is the volume of this wedge?
   Answer: $\frac{2}{3}$ The volume $V$ is given by the triple integral $\iiint_E 1 \, dV$. The region $E$ is described by $x^2+y^2 \le 1$, $y \ge 0$, and $0 \le z \le y$. The projection of the solid onto the $xy$-plane is the upper semi-disk $D$. The volume integral is $V = \iint_D \left( \int_0^y dz \right) dA = \iint_D y \, dA$. We can set this up as an iterated integral in Cartesian coordinates: $V = \int_0^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} y \, dx \, dy$. The inner integral gives $y[x]_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} = 2y\sqrt{1-y^2}$. The outer integral is $\int_0^1 2y\sqrt{1-y^2} \, dy$. Using the substitution $u=1-y^2$, $du=-2y \, dy$, we get $\int_1^0 -\sqrt{u} \, du = \int_0^1 u^{1/2} \, du = [\frac{2}{3}u^{3/2}]_0^1 = \frac{2}{3}$.
7. A solid region is bounded above by the plane $z = 4y$ and below by the paraboloid $z = x^2 + y^2$. Which of the following iterated integrals represents the volume of this solid?
   Answer: $\int_{-2}^2 \int_{2-\sqrt{4-x^2}}^{2+\sqrt{4-x^2}} \int_{x^2+y^2}^{4y} dz \, dy \, dx$ The intersection of the two surfaces $z = 4y$ and $z = x^2 + y^2$ occurs when $4y = x^2+y^2$, which can be rewritten as $x^2 + y^2 - 4y = 0$, or $x^2 + (y-2)^2 = 4$. This is a circle of radius 2 centered at $(0,2)$ in the $xy$-plane, which is the projection of the solid. The solid is bounded below by $z=x^2+y^2$ and above by $z=4y$, so the innermost integral is $\int_{x^2+y^2}^{4y} dz$. For the bounds of integration in the $xy$-plane, $x$ ranges from $-2$ to $2$. For a fixed $x$, $y$ is bounded by the circle $x^2+(y-2)^2=4$, so $(y-2)^2 = 4-x^2$, which means $y-2 = \pm\sqrt{4-x^2}$, or $y = 2 \pm \sqrt{4-x^2}$. This gives the bounds for the middle integral.
8. A student sets up a triple integral to find the mass of a solid $E$ with density function $\rho(x,y,z)$. The solid $E$ is the pyramid with a square base on the $xy$-plane defined by $-1 \le x \le 1$, $-1 \le y \le 1$, and a vertex at $(0,0,2)$. The student's setup is: $\int_{-1}^1 \int_{-1}^1 \int_0^{2-2\sqrt{x^2+y^2}} \rho(x,y,z) \, dz \, dy \, dx$ What is the error in this setup?
   Answer: The bounds for $z$ describe a cone, not a pyramid. The equation for the surface of the pyramid must be linear. The pyramid consists of four triangular faces. For example, the face in the region $x>0, y>0$ and connecting the vertex $(0,0,2)$ to the edge from $(1,0,0)$ to $(0,1,0)$ would lie on the plane $z = 2-2x-2y$. The upper bound for $z$ would be a piecewise function of four such planes. The expression $z=2-2\sqrt{x^2+y^2}$ describes a cone with its vertex at $(0,0,2)$ and its base as the circle $x^2+y^2=1$ on the $xy$-plane. The student has incorrectly modeled the slanted faces of the pyramid with a conical surface.
9. Which of the following iterated integrals represents the volume of the solid region inside the sphere $x^2+y^2+z^2=4$ and inside the cylinder $x^2+y^2=1$?
   Answer: $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} dz \, dy \, dx$ When setting up triple integrals for volumes, you need to carefully identify the region of integration by finding where the given surfaces intersect and which surface provides the boundary at each point. The region is bounded by both a sphere $x^2+y^2+z^2=4$ (radius 2) and a cylinder $x^2+y^2=1$ (radius 1). Since the cylinder has a smaller radius than the sphere, the cylinder sits entirely inside the sphere, creating a cylindrical region that extends vertically until it hits the sphere's surface. For the correct setup, start with the $xy$-projection. Since we want the region inside the cylinder $x^2+y^2=1$, we integrate $x$ from $-1$ to $1$, and for each $x$, $y$ ranges from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$. For the $z$-limits, since we're inside both surfaces, $z$ extends from the bottom of the sphere to the top: $-\sqrt{4-x^2-y^2}$ to $\sqrt{4-x^2-y^2}$. Option A uses $\sqrt{1-x^2-y^2}$ for the $z$-limits, which represents the cylinder's boundary—but cylinders don't have $z$-boundaries. Option B incorrectly uses the sphere's $xy$-projection ($x$ from $-2$ to $2$) instead of the cylinder's. Option C only integrates the upper half of the region by using $z$ from $0$ to $\sqrt{4-x^2-y^2}$, giving half the volume. Option D correctly captures the entire cylindrical region bounded by the sphere. Study tip: Always identify which surface provides each boundary. The intersection region's $xy$-projection comes from the smaller radius, while $z$-limits come from surfaces that actually bound $z$ (like spheres, not cylinders).
10. Let $E$ be the solid region bounded by the surfaces $z = \sqrt{x^2+y^2}$ and $z = 6 - (x^2+y^2)$. The volume of $E$ is given by the triple integral $\iiint_E 1 \, dV$. If this integral is set up as $\int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \int_{\sqrt{x^2+y^2}}^{6-(x^2+y^2)} dz \, dy \, dx$, what is the value of $a$?
    Answer: $2$ When setting up triple integrals for volumes bounded by surfaces, you need to identify where the surfaces intersect to determine the limits of integration. The key insight is that the outer limits of integration correspond to the projection of the region onto the xy-plane. To find this projection, you need to determine where the two surfaces $z = \sqrt{x^2+y^2}$ and $z = 6 - (x^2+y^2)$ intersect. Setting them equal: $\sqrt{x^2+y^2} = 6 - (x^2+y^2)$ Let $r = \sqrt{x^2+y^2}$, so $r = 6 - r^2$. Rearranging: $r^2 + r - 6 = 0$, which factors as $(r+3)(r-2) = 0$. Since $r \geq 0$, we have $r = 2$, meaning $\sqrt{x^2+y^2} = 2$. This intersection occurs on the circle $x^2 + y^2 = 4\$, which has radius 2. The projection of region E onto the xy-plane is therefore the disk of radius 2 centered at the origin. In Cartesian coordinates, this means$x$ranges from$-2$to$2$, so$a = 2$. Looking at the wrong answers: A) 3 might come from incorrectly solving the intersection equation or confusing it with another parameter. B)$\sqrt{5}$and C)$\sqrt{6}$likely result from algebraic errors when solving$$r^2$ + r - 6 = 0$$ or from misunderstanding which surfaces bound the region. Study tip: Always find intersection curves first when setting up triple integrals with multiple bounding surfaces. The intersection determines the boundary of your region's projection, which gives you the outer integration limits.
11. A solid occupies the region in the first octant bounded by the coordinate planes and the plane $x+y+z=1$. The density of the solid is given by the function $\rho(x,y,z) = y$. What is the total mass of the solid?
    Answer: $\frac{1}{24}$ The mass $M$ is the triple integral of the density function over the region $E$. The region is a tetrahedron defined by $x \ge 0, y \ge 0, z \ge 0,$ and $x+y+z \le 1$. We set up the integral as $M = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} y \, dz \, dy \, dx$. The innermost integral (with respect to $z$) is $\int_0^{1-x-y} y \, dz = y(1-x-y) = y - xy - y^2$. The middle integral (with respect to $y$) is $\int_0^{1-x} (y-xy-y^2) \, dy = [\frac{y^2}{2} - \frac{xy^2}{2} - \frac{y^3}{3}]_0^{1-x} = \frac{(1-x)^2}{2} - \frac{x(1-x)^2}{2} - \frac{(1-x)^3}{3} = \frac{(1-x)^3}{2} - \frac{(1-x)^3}{3} = \frac{(1-x)^3}{6}$. The outer integral (with respect to $x$) is $\int_0^1 \frac{(1-x)^3}{6} \, dx = [-\frac{(1-x)^4}{24}]_0^1 = 0 - (-\frac{1}{24}) = \frac{1}{24}$.
12. Evaluate the iterated integral: $\int_0^1 \int_{\sqrt[3]{z}}^1 \int_0^{\ln(3)} \frac{\pi e^x \sin(y^2)}{y^2} \, dx \, dy \, dz$
    Answer: $\pi(1 - \cos(1))$ The integral with respect to $y$ is problematic. First, integrate with respect to $x$: $\int_0^{\ln(3)} \pi e^x dx = [\pi e^x]_0^{\ln 3} = \pi(3-1) = 2\pi$. The problem reduces to $2\pi \int_0^1 \int_{\sqrt[3]{z}}^1 \frac{\sin(y^2)}{y^2} \, dy \, dz$. Now, we change the order of integration for $y$ and $z$. The region is $0 \le z \le 1$ and $\sqrt[3]{z} \le y \le 1$. This is equivalent to $0 \le y \le 1$ and $0 \le z \le y^3$. The integral becomes $2\pi \int_0^1 \int_0^{y^3} \frac{\sin(y^2)}{y^2} \, dz \, dy = 2\pi \int_0^1 \frac{\sin(y^2)}{y^2} [z]_0^{y^3} \, dy = 2\pi \int_0^1 y \sin(y^2) \, dy$. Using substitution $u=y^2$, $du=2y \, dy$, we get $\pi \int_0^1 \sin(u) \, du = \pi [-\cos(u)]_0^1 = \pi(-\cos(1) - (-1)) = \pi(1-\cos(1))$.
13. Consider the solid region $W$ defined by $x^2 + y^2 \leq 9$, $0 \leq z \leq x^2 + y^2$. If we want to evaluate $\iiint_W z \, dV$ using the order $dx \, dz \, dy$, which of the following setups is correct?
    Answer: $\int_{-3}^{3} \int_{y^2}^{9} \int_{-\sqrt{z-y^2}}^{\sqrt{z-y^2}} z \, dx \, dz \, dy$ The region is bounded by $x^2 + y^2 \leq 9$ and $0 \leq z \leq x^2 + y^2$. For order $dx \, dz \, dy$: $y$ ranges from $-3$ to $3$. For fixed $y$, we need $x^2 + y^2 \leq 9$ and $z \leq x^2 + y^2$, so $x^2 \geq z - y^2$. Since $x^2 + y^2 \leq 9$, we have $z \leq 9$. But we also need $z \geq x^2 + y^2 \geq y^2$ (since $x^2 \geq 0$), and $z \geq 0$. So $z$ ranges from $y^2$ to $9$. For fixed $y$ and $z$, we need $x^2 = z - y^2$ (approximately), so $x$ ranges from $-\sqrt{z-y^2}$ to $\sqrt{z-y^2}$, provided $z \geq y^2$. Choice A has wrong $z$-bounds. Choice B has wrong $x$-bounds. Choice D allows $z < y^2$, making $\sqrt{z-y^2}$ undefined.
14. Consider the solid $S$ bounded by $z = x^2 + y^2$ and $z = 8 - x^2 - y^2$. Which of the following correctly describes a necessary condition for the limits of integration when evaluating $\iiint_S f(x,y,z) \, dV$ in the order $dz \, dx \, dy$?
    Answer: The $y$-limits must satisfy $|y| \leq 2$ and the $x$-limits must depend on $y$ as $|x| \leq \sqrt{4-y^2}$ The solids intersect when $x^2 + y^2 = 8 - x^2 - y^2$, giving $2(x^2 + y^2) = 8$, so $x^2 + y^2 = 4$. The projection onto the $xy$-plane is the disk $x^2 + y^2 \leq 4$. For the order $dz \, dx \, dy$, we integrate $y$ outermost, so $-2 \leq y \leq 2$. For fixed $y$, $x$ ranges over $-\sqrt{4-y^2} \leq x \leq \sqrt{4-y^2}$ (the cross-section of the disk). For fixed $(x,y)$, $z$ ranges from $x^2 + y^2$ to $8 - x^2 - y^2$. Choice B uses the wrong radius ($2\sqrt{2}$ instead of $2$). Choice C incorrectly makes $x$-limits independent of $y$. Choice D incorrectly suggests the limits depend on the integrand function.

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