📝 Volume Calculation by Triple Integral (14 MCQs)

📖 Explanation: The student's setup has a critical error in the $z$-limits. For a point with cylindrical radius $r$, the sphere gives $z = \pm \sqrt{4 - r^2}$. However, the region outside the cylinder $r=1$ but inside the sphere means for a given $r$, $z$ goes from the lower sphere to the upper sphere, which is correctly represented as $\pm \sqrt{4 - r^2}$. The error is that the $r$ limits are correct (1 to 2), but the student forgot that for a point at radius $r$, the cylinder $r=1$ is a boundary in the $xy$-plane, not a surface in 3D. The setup is actually correct. This is a trick question to test if the student understands that the cylinder's equation $x^2+y^2=1$ translates directly to $r=1$ in cylindrical coordinates, and the sphere to $r^2+z^2=4$, making the $z$-limits dependent on $r$. The correct answer is that the setup is correct, which is option A. The misconception is that students might think the cylinder affects the $z$-limits, but it only affects the $r$-limits.

📖 Explanation: The region is bounded by the cylinder $x^2 + z^2 = 1$ which extends along the $y$-axis. The first octant implies $x \ge 0, y \ge 0, z \ge 0$. The plane $y=4$ gives the upper bound for $y$. The projection on the $xz$-plane is the quarter circle $x^2+z^2 \le 1$ with $x \ge 0, z \ge 0$. The student's integral $\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{0}^{4} dy \, dz \, dx$ has the limits for $z$ from 0 to $\sqrt{1-x^2}$, and $x$ from 0 to 1. This correctly describes the quarter circle. The order $dy \, dz \, dx$ means for a fixed $x$, $z$ varies, and for each $x,z$, $y$ varies from 0 to 4. This is a correct setup. The answer is A. The distractor D is a common error where students incorrectly place the variable in the limit.

📖 Explanation: The tetrahedron's intercepts are at $x=6, y=4, z=3$. The volume can be calculated as $\int_{0}^{6} \int_{0}^{4 - \frac{2}{3}x} \int_{0}^{3 - \frac{1}{2}x - \frac{3}{4}y} dz \, dy \, dx$. The inner integral gives $3 - \frac{1}{2}x - \frac{3}{4}y$. The next integral $\int_{0}^{4 - \frac{2}{3}x} (3 - \frac{1}{2}x - \frac{3}{4}y) dy$ evaluates to $\frac{(4 - \frac{2}{3}x)^2}{2} \cdot \frac{3}{4} = \frac{3}{8}(4 - \frac{2}{3}x)^2$. The final integral $\int_{0}^{6} \frac{3}{8}(4 - \frac{2}{3}x)^2 dx$ equals 12. The volume of a tetrahedron with intercepts $a, b, c$ is $abc/6 = 6*4*3/6 = 12$. Option B is correct.

📖 Explanation: The integral is in spherical coordinates $(\rho, \phi, \theta)$. The limits are $\rho: 0 \to 2$, $\phi: 0 \to \pi$, and $\theta: 0 \to 2\pi$. This describes the entire sphere of radius 2 centered at the origin. The Jacobian for spherical coordinates is $\rho^2 \sin\phi$. The volume of a sphere of radius $R$ is $\frac{4}{3}\pi R^3 = \frac{32}{3}\pi$. The integral evaluates to that. Option A is the direct recall. The other options are common solids but with different coordinate systems or limits.

📖 Explanation: The region is bounded below by the paraboloid $z = r^2$ and above by the plane $z = 4$ in cylindrical coordinates. The intersection is $r^2 = 4 \Rightarrow r = 2$. Thus $r$ goes from 0 to 2. The volume element is $r \, dz \, dr \, d\theta$. Option A has the correct limits and the Jacobian $r$. Option B has the wrong upper limit for $z$ (it should be 4, not $4-r^2$). Option C has the wrong upper limit for $r$ (0 to 4). Option D is missing the Jacobian $r$. This tests the student's ability to convert from Cartesian to cylindrical and correctly apply the Jacobian and limits.

📖 Explanation: The region is 'inside the sphere $\rho=2$' and 'above the cone $\phi = \pi/4$'. In spherical coordinates, the cone $\phi = \pi/4$ opens upwards. The region above this cone means the polar angle $\phi$ is measured from the positive z-axis. 'Above' the cone means $\phi$ is smaller than $\pi/4$ (closer to the z-axis). However, the region inside the sphere and above the cone in the first quadrant is actually for $\phi$ from 0 to $\pi/4$ for the upper part, but to get the full solid inside the sphere and above the cone (which extends in all directions), $\phi$ goes from 0 to $\pi/4$ for the upper cone, and from $3\pi/4$ to $\pi$ for the lower cone? No, the region 'above the cone' typically means $z \ge \sqrt{x^2+y^2}$, which in spherical is $\cos\phi \ge \sin\phi \Rightarrow \phi \le \pi/4$. The student's integral has $\phi$ from 0 to $\pi/4$, which correctly describes the region inside the sphere and above the cone. So the setup is correct. The correct answer is A. The distractor D is a common error where students think above the cone means $\phi > \pi/4$ (which is below the cone).

📖 Explanation: The solid is bounded by two cylinders: $x^2+y^2=1$ (independent of z) and $x^2+z^2=1$ (independent of y). In the first octant, $x, y, z \ge 0$. The projection on the $xy$-plane is the quarter circle $x^2+y^2 \le 1$, so for a given $x$, $y$ goes from 0 to $\sqrt{1-x^2}$. The cylinder $x^2+z^2=1$ bounds $z$, so for a given $x$, $z$ goes from 0 to $\sqrt{1-x^2}$. Thus the volume is $\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2}} dz \, dy \, dx$. Option B matches this. Option A has the order of $dy$ and $dz$ swapped but the same limits; however, in a triple integral, the order matters for the limits, but here the limits for $y$ and $z$ are identical functions of $x$, so $\int dz \, dy$ and $\int dy \, dz$ are equivalent in value. The question asks for 'correctly represents', and B is the standard form. Option C and D have incorrect limits based on the other variables.

📖 Explanation: The inner integral limit $z = \sqrt{4-x^2-y^2}$ and the middle limit $y = \sqrt{4-x^2}$ and the outer limit $x=2$ describe the first octant of a sphere. The equation $x^2+y^2+z^2=4$ gives $z = \sqrt{4-x^2-y^2}$. The projection on the $xy$-plane is $x^2+y^2 \le 4$, and the first octant restricts $x,y,z \ge 0$. The integral correctly computes the volume of the first octant of a sphere of radius 2. The volume of this region is $\frac{1}{8} \cdot \frac{4}{3}\pi (2)^3 = \frac{4}{3}\pi$. Option A is correct. The other options are common solids that would have different integrands or limits.

📖 Explanation: The two surfaces are $z = r^2$ (paraboloid opening up) and $z = 2 - r^2$ (paraboloid opening down). Their intersection is $r^2 = 2 - r^2 \Rightarrow 2r^2 = 2 \Rightarrow r=1$. The region between them has $z$ from $r^2$ to $2-r^2$, $r$ from 0 to 1, and $\theta$ from 0 to $2\pi$. The volume element is $r \, dz \, dr \, d\theta$. Option A has the correct limits and Jacobian. Option B has the wrong upper limit for $r$ ($\sqrt{2}$ is the radius where the paraboloid $z=2-r^2$ meets the plane $z=0$, not the intersection). Option C is missing the Jacobian $r$. Option D has the wrong lower limit for $z$ (it should be $r^2$, not 0) and the wrong upper limit for $r$.

📖 Explanation: The region is bounded by the paraboloid $z = x^2+y^2$ and the plane $z=4$. The intersection is the circle $x^2+y^2=4$ in the plane $z=4$. The integral is in cylindrical coordinates (but written in Cartesian). The volume is $\int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{4} r \, dz \, dr \, d\theta$. The inner integral gives $4 - r^2$. The next integral $\int_{0}^{2} (4r - r^3) dr = [2r^2 - r^4/4]_0^2 = 8 - 4 = 4$. Then $\int_{0}^{2\pi} 4 d\theta = 8\pi$. The student's claim of $16\pi$ is double the correct volume. The error might be from forgetting the $1/2$ or miscomputing the area of the circle. The correct volume is $8\pi$. Option B is correct.

📖 Explanation: The sphere $x^2+y^2+z^2 = 4z$ can be rewritten as $x^2+y^2+(z-2)^2 = 4$. In cylindrical coordinates, this is $r^2 + (z-2)^2 = 4$, so $z = 2 \pm \sqrt{4-r^2}$. The paraboloid is $z = r^2$. The region inside the sphere and above the paraboloid means $z$ is between $r^2$ (paraboloid) and $2+\sqrt{4-r^2}$ (upper half of sphere). The intersection of the paraboloid and sphere is $r^2 = 2 + \sqrt{4-r^2}$. Solving: let $u = r^2$, then $u = 2 + \sqrt{4-u} \Rightarrow u-2 = \sqrt{4-u} \Rightarrow (u-2)^2 = 4-u \Rightarrow u^2 -4u+4 = 4-u \Rightarrow u^2 -3u = 0 \Rightarrow u=0$ or $u=3$. The non-zero intersection is $r = \sqrt{3}$. So $r$ goes from 0 to $\sqrt{3}$. The correct integral is $\int_{0}^{2\pi} \int_{0}^{\sqrt{3}} \int_{r^2}^{2+\sqrt{4-r^2}} r \, dz \, dr \, d\theta$. Option A matches this. Option B has $r$ from 0 to 2, which is the sphere's radius, not the intersection. Option C has the wrong lower limit for $z$ (0 instead of $r^2$). Option D uses the lower half of the sphere $2-\sqrt{4-r^2}$, which is below the paraboloid. This is a challenging problem requiring algebraic manipulation to find the intersection.

📖 Explanation: The solid is a hemisphere of radius $R$ with a cylindrical hole of radius $a$ drilled through its center along the axis. The hemisphere is assumed to be the upper half of a sphere ($z \ge 0$) with the hole along the z-axis. The volume is the region inside the hemisphere and outside the cylinder $r=a$. In cylindrical coordinates, $r$ goes from $a$ to $R$, $\theta$ from 0 to $2\pi$, and $z$ from 0 to $\sqrt{R^2-r^2}$. The integral is correct. The subtraction method would be: Volume of hemisphere ($(2/3)\pi R^3$) minus volume of cylinder of radius $a$ and height $R$ (since the cylinder's height is the radius $R$ of the hemisphere) = $\frac{2}{3}\pi R^3 - \pi a^2 R$. The integral $\int_{0}^{2\pi} \int_{a}^{R} \int_{0}^{\sqrt{R^2-r^2}} r \, dz \, dr \, d\theta$ evaluates to $\int_{0}^{2\pi} \int_{a}^{R} r\sqrt{R^2-r^2} dr d\theta = 2\pi [-\frac{1}{3}(R^2-r^2)^{3/2}]_a^R = \frac{2\pi}{3}(R^2-a^2)^{3/2}$. Wait, this is not the same as the subtraction method unless $a=0$. The error is that the height of the cylinder drilled is not $R$; the cylinder is bounded by the hemisphere, so its height varies with $r$. The subtraction method would be complicated. The integral is correct for the solid described. Option A is correct. The other options represent common misconceptions about the limits and the subtraction method.

📖 Explanation: The integral $V = \int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} dz \, dy \, dx$ describes the volume common to two cylinders $x^2+y^2 \le a^2$ and $x^2+z^2 \le a^2$. The inner integral gives $2\sqrt{a^2-x^2}$. The next integral \int_{-a}^{a} \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} 2\sqrt{a^2-x^2} dy \, dx = \int_{-a}^{a} 4(a^2-x^2) dx = 4[a^2 x - x^3/3]_{-a}^{a} = 4( (a^3 - a^3/3) - (-a^3 + a^3/3) ) = 4( (2a^3/3) - (-2a^3/3) ) = 4(4a^3/3) = 16a^3/3. For \(a=1, the volume is $16/3$. Option A is correct. This is a classic problem (Steinmetz solid) and requires understanding of the geometry and the limits. The distractors are common incorrect results from missing factors or misapplying the limits.

📖 Explanation: The region is a tetrahedron bounded by the coordinate planes and the plane $x+y+z=1$. The correct integral is $\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} dz \, dy \, dx$. The inner integral gives $1-x-y$. The next integral $\int_{0}^{1-x} (1-x-y) dy = [(1-x)y - y^2/2]_0^{1-x} = (1-x)^2 - (1-x)^2/2 = (1-x)^2/2$. Then $\int_{0}^{1} (1-x)^2/2 dx = [-(1-x)^3/6]_0^1 = 1/6$. The first student is correct. The second student's integral $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1-x-y} dz \, dy \, dx$ has the wrong limits for $y$ (it should depend on $x$). This would incorrectly integrate over a square region where $x+y$ can exceed 1, making the integrand negative. The volume is $\frac{1}{6}$. Option A is correct. This tests the understanding of the correct limits for a tetrahedron.