Triple Integrals in Cylindrical - Calculus 3

1. Consider the integral $\int_0^{2\pi} \int_0^2 \int_0^{4-r^2} r^3 \cos\theta \, dz \, dr \, d\theta$. After performing the integration with respect to $z$, which expression correctly represents the remaining double integral?
   Answer: $\int_0^{2\pi} \int_0^2 r^3(4-r^2)\cos\theta \, dr \, d\theta$ When you encounter a triple integral in cylindrical coordinates, you'll typically integrate from the innermost variable outward. Here, you're integrating with respect to $z$ first, then $r$, then $\theta$. To integrate $\int_0^{4-r^2} r^3 \cos\theta \, dz$, notice that the integrand $r^3 \cos\theta$ doesn't contain $z$, so it acts as a constant with respect to the $z$-integration. When you integrate a constant over an interval, you multiply that constant by the length of the interval. The integration becomes: $\int_0^{4-r^2} r^3 \cos\theta \, dz = r^3 \cos\theta \int_0^{4-r^2} dz = r^3 \cos\theta \cdot z \Big|_0^{4-r^2} = r^3 \cos\theta \cdot (4-r^2)$ This gives you the double integral $\int_0^{2\pi} \int_0^2 r^3(4-r^2)\cos\theta \, dr \, d\theta$, which is answer D. Let's examine why the other options are incorrect: Answer A drops the $r^3$ term entirely, which would only happen if the original integrand were just $\cos\theta$. Answer B incorrectly changes $r^3$ to $r^2$, suggesting a misunderstanding of how constants behave during integration. Answer C incorrectly changes $\cos\theta$ to $\sin\theta$, which has no basis in the integration process. Study tip: When integrating with respect to one variable, treat all other variables as constants. The key is identifying what depends on your integration variable and what doesn't.
2. The integral $\int_0^{\pi} \int_0^{2\cos\theta} \int_0^{r\sin\theta} r^2 \, dz \, dr \, d\theta$ represents integration over which type of region?
   Answer: A region bounded by a circular cylinder, a plane, and the xy-plane When you encounter a triple integral in cylindrical coordinates, you need to analyze the bounds to visualize the 3D region. The integral has the form $\int \int \int r^2 \, dz \, dr \, d\theta$, where the $r^2$ factor (not just $r$) and the order $dz \, dr \, d\theta$ indicate cylindrical coordinates. Let's examine each bound systematically. The $\theta$ ranges from $0$ to $\pi$ (a semicircle), $r$ goes from $0$ to $2\cos\theta$, and $z$ extends from $0$ to $r\sin\theta$. The key insight is recognizing that $r = 2\cos\theta$ describes a circle in the $xy$-plane. Converting to Cartesian coordinates: $r = 2\cos\theta$ becomes $r^2 = 2r\cos\theta$, which gives $x^2 + y^2 = 2x$, or $(x-1)^2 + y^2 = 1$. This is a circle centered at $(1,0)$ with radius $1$. The upper bound $z = r\sin\theta$ represents a plane, since in Cartesian coordinates this becomes $z = y$. The lower bound $z = 0$ is simply the $xy$-plane. Therefore, the region is bounded by a circular cylinder (extending the circle $(x-1)^2 + y^2 = 1$ vertically), the plane $z = y$, and the $xy$-plane, making B correct. A is wrong because $r = 2\cos\theta$ creates a circle, not a parabola. C is incorrect since the cross-section is circular, not elliptical. D fails because a circle isn't a hyperbola. Study tip: When analyzing cylindrical coordinate bounds, always convert $r = f(\theta)$ to Cartesian form to identify the curve type—this immediately reveals the cylinder's cross-section.
3. A solid $E$ is bounded by the paraboloid $z = x^2 + y^2$ and the plane $z = 9$. To evaluate $\iiint_E z \, dV$ using cylindrical coordinates, which of the following represents the correct setup?
   Answer: $\int_0^{2\pi} \int_0^3 \int_{r^2}^9 z \cdot r \, dz \, dr \, d\theta$ In cylindrical coordinates, $x^2 + y^2 = r^2$, so the paraboloid becomes $z = r^2$. The solid extends from $z = r^2$ to $z = 9$, and the projection onto the xy-plane is the disk $r^2 \leq 9$, so $0 \leq r \leq 3$. The Jacobian for cylindrical coordinates is $r$, giving the integrand $z \cdot r$. Choice A correctly sets up these bounds with the proper Jacobian.
4. A solid region is bounded by $z = 0$, $z = h$, and $x^2 + y^2 = az$ where $a > 0$ and $h > 0$. The triple integral $\iiint_E \rho \, dV$ where $\rho = \sqrt{x^2 + y^2 + z^2}$ represents the total mass if $\rho$ is the density. In cylindrical coordinates, which integrand is correct?
   Answer: $\sqrt{r^2 + z^2} \cdot r$ When converting triple integrals to cylindrical coordinates, you need to transform both the integrand and include the Jacobian determinant. This is a classic coordinate transformation problem that tests your understanding of how volume elements change between coordinate systems. The correct integrand is $\sqrt{r^2 + z^2} \cdot r$. Here's why: In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and $z = z$. The density function $\rho = \sqrt{x^2 + y^2 + z^2}$ becomes $\sqrt{r^2 + z^2}$ since $x^2 + y^2 = r^2$. Crucially, the volume element $dV = dx\,dy\,dz$ transforms to $r\,dr\,d\theta\,dz$, where the factor $r$ is the Jacobian. Therefore, the integrand must include both the transformed density and this Jacobian factor. Option A ($\sqrt{r^2 + z^2}$) correctly transforms the density but omits the essential Jacobian factor $r$. Option B ($\sqrt{r + z^2} \cdot r$) includes the Jacobian but incorrectly transforms the density—it should be $r^2$, not just $r$, under the square root. Option D ($r\sqrt{r^2 + z^2} \cdot z$) includes an erroneous factor of $z$ that has no basis in the coordinate transformation. Study tip: Always remember that cylindrical coordinate transformations require the Jacobian factor $r$ to be included in your integrand. When you see $x^2 + y^2$ in the original problem, it always becomes $r^2$ in cylindrical coordinates.
5. Consider the region $E$ where $1 \leq x^2 + y^2 \leq 4$ and $0 \leq z \leq xy$. To express $\iiint_E f(x,y,z) \, dV$ in cylindrical coordinates, what constraint must be placed on $\theta$?
   Answer: $0 \leq \theta \leq \frac{\pi}{2}$ and $\pi \leq \theta \leq \frac{3\pi}{2}$ When converting triple integrals to cylindrical coordinates, you must carefully analyze all constraints to determine the valid range for each variable. The key insight here is recognizing when the $z$-bounds create additional restrictions on the angular variable $\theta$. The constraint $0 \leq z \leq xy$ is crucial. Since $z \geq 0$, we need $xy \geq 0$. In cylindrical coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, so $xy = r^2\cos\theta\sin\theta$. Since $r^2 > 0$ (from $1 \leq x^2 + y^2 \leq 4$), we need $\cos\theta\sin\theta \geq 0$. The product $\cos\theta\sin\theta$ is non-negative when both trigonometric functions have the same sign. This occurs in Quadrant I ($0 \leq \theta \leq \frac{\pi}{2}$) where both are positive, and in Quadrant III ($\pi \leq \theta \leq \frac{3\pi}{2}$) where both are negative. Therefore, answer C is correct. Answer A ($0 \leq \theta \leq 2\pi$) includes all quadrants, but $\cos\theta\sin\theta < 0$ in Quadrants II and IV, making $xy < 0$ and violating $z \geq 0$. Answer B ($0 \leq \theta \leq \pi$) includes Quadrant II where $\cos\theta < 0$ and $\sin\theta > 0$. Answer D gives incorrect specific angles that don't correspond to the quadrants where $\cos\theta\sin\theta \geq 0$. Study tip: When converting to cylindrical coordinates, always check if constraints involving products like $xy$ create sign restrictions that limit $\theta$ to specific quadrants.
6. A region $E$ is defined by $x^2 + y^2 \leq 4$, $x^2 + y^2 \leq z \leq 6$. To find the volume of $E$ using cylindrical coordinates, what should be the order of integration and bounds?
   Answer: $\int_0^{2\pi} \int_0^2 \int_{r^2}^6 r \, dz \, dr \, d\theta$ When setting up triple integrals in cylindrical coordinates, you need to carefully analyze the region's geometry to determine the correct order of integration and bounds. The region $E$ is defined by $x^2 + y^2 \leq 4$ (a cylinder of radius 2) and $x^2 + y^2 \leq z \leq 6$. In cylindrical coordinates, $x^2 + y^2 = r^2$, so the region becomes $r^2 \leq z \leq 6$ with $0 \leq r \leq 2$. The correct approach uses order $dz \, dr \, d\theta$ because for each fixed $r$ and $\theta$, $z$ varies from the bottom surface $z = r^2$ to the top at $z = 6$. The radial coordinate $r$ goes from 0 to 2 (the cylinder's radius), and $\theta$ completes a full revolution from 0 to $2\pi$. This gives us answer D: $\int_0^{2\pi} \int_0^2 \int_{r^2}^6 r \, dz \, dr \, d\theta$. Option A incorrectly integrates $r$ from 0 to 2 regardless of $z$, ignoring that the region's cross-section changes with height. Option B uses $\min(2,\sqrt{z})$ which would apply if we integrated $dr$ last, but this creates unnecessary complexity and uses the wrong order. Option C has $z$ going from 0 to 6, but the region doesn't extend below $z = r^2$. Study tip: Always sketch the region first, then choose the integration order that makes the bounds simplest. In cylindrical coordinates, if the region has circular symmetry with varying height, often $dz \, dr \, d\theta$ works best.
7. For the triple integral $\int_0^{2\pi} \int_0^{\sqrt{2}} \int_{r^2}^{2-r^2} f(r,\theta,z) \, r \, dz \, dr \, d\theta$, at what value of $r$ do the upper and lower bounds of integration for $z$ intersect?
   Answer: $r = 1$ When you encounter a triple integral with variable bounds in cylindrical coordinates, pay attention to where the integration limits intersect—this often reveals important geometric information about the region. To find where the upper and lower bounds of $z$ intersect, you need to solve for when they're equal. The bounds are $z = r^2$ (lower) and $z = 2 - r^2$ (upper). Setting them equal: $r^2 = 2 - r^2$. Solving this equation: $2r^2 = 2$, so $r^2 = 1$, which gives $r = 1$ (taking the positive value since $r \geq 0$ in cylindrical coordinates). You can verify this makes sense geometrically. At $r = 1$: the lower bound gives $z = 1^2 = 1$, and the upper bound gives $z = 2 - 1^2 = 1$. Both bounds equal 1, confirming they intersect at this point. Let's check why the other options fail: Choice (A) $r = \sqrt{2}$ gives $z = 2$ for the lower bound and $z = 0$ for the upper bound—they don't intersect. Choice (C) $r = \frac{\sqrt{2}}{2}$ gives $z = \frac{1}{2}$ and $z = \frac{3}{2}$—different values, so no intersection. Choice (D) $r = 2$ exceeds the integration limit $r \leq \sqrt{2}$, making it outside the domain entirely. Study tip: When finding intersection points of bounds in multiple integrals, always set the expressions equal and solve algebraically. Then verify by substituting back into both original expressions to ensure they give the same value.
8. The triple integral $\int_0^{2\pi} \int_0^4 \int_{\sqrt{r}}^{8-r} f(r,\theta,z) \, r \, dz \, dr \, d\theta$ represents integration over a region $E$. Which statement best describes the boundary surfaces of this region?
   Answer: The region is bounded below by the surface $z = \sqrt{x^2 + y^2}$ and above by the plane $z = 8 - \sqrt{x^2 + y^2}$ From the limits of integration, $z$ ranges from $\sqrt{r}$ to $8-r$. In cylindrical coordinates, $r = \sqrt{x^2 + y^2}$, so the lower boundary is $z = \sqrt{\sqrt{x^2 + y^2}} = \sqrt{x^2 + y^2}$ and the upper boundary is $z = 8 - \sqrt{x^2 + y^2}$. Choice B incorrectly converts $\sqrt{r}$ to $z^2 = x^2 + y^2$. Choice C mistakes the lower boundary. Choice D has both errors.

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