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Evaluate the following integrals in cylindrical coordinates. $$\int_{-4}^{4} \int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{4} d z d y d x$$

Short Answer

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Question: Calculate the value of the integral $\int_{-4}^{4} \int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{4} dz dy dx$ in cylindrical coordinates. Answer: The value of the integral in cylindrical coordinates is $8\pi\left(32 - \frac{64}{3}\right)$.

Step by step solution

01

Find the Jacobian of the transformation from Cartesian to cylindrical coordinates

In cylindrical coordinates, we use the transformation equations as follows: $$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ $$z = z$$ Calculate the Jacobian determinant of this transformation: $$J = \frac{\partial(x, y, z)}{\partial(r, \theta, z)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{vmatrix} = \begin{vmatrix} \cos(\theta) & -r\sin(\theta) & 0 \\ \sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1\end{vmatrix} = r\cos^2(\theta) + r\sin^2(\theta) = r$$
02

Transform the given integral into cylindrical coordinates

Now, substitute the conversion formulas and the Jacobian into the given integral: $$\int_{-4}^{4} \int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{4} dz dy dx = \int_{0}^{2\pi} \int_{0}^{4} \int_{r}^{4} r dz dr d\theta$$
03

Evaluate the integral

Solve the integral step by step: 1. Integrate over z: $$\int_{r}^{4} r dz = rz\Big|_{r}^{4} = 4r - r^2$$ 2. Integrate over r: $$\int_{0}^{4} (4r - r^2) dr = 2r^2 - \frac{r^3}{3}\Big|_{0}^{4} = 32 - \frac{64}{3} $$ 3. Integrate over θ: $$\int_{0}^{2\pi} \left(32 - \frac{64}{3}\right) d\theta = \left(32 - \frac{64}{3}\right)2\pi\theta\Big|_{0}^{2\pi} = \left(32 - \frac{64}{3}\right)2\pi(2\pi) = 8\pi\left(32 - \frac{64}{3}\right)$$ The final result of the given integral in cylindrical coordinates is: $$\int_{-4}^{4} \int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}} \int_{\sqrt{x^{2}+y^{2}}}^{4} dz dy dx = 8\pi\left(32 - \frac{64}{3}\right)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus. It is used to find areas, volumes, and other quantities that accumulate. In this exercise, we're shifting from the Cartesian coordinate system to the cylindrical coordinate system for ease of computation.
Integrating in different coordinate systems can simplify problems significantly, especially when dealing with symmetric shapes in three dimensions. For instance, using cylindrical coordinates when dealing with cylinders or circular regions is advantageous.

This exercise starts with a triple integral in Cartesian coordinates. It is then transformed into cylindrical coordinates. This involves integrating with respect to the height of the cylinder (z), the radius (r), and the angle (θ). Each integral is evaluated step by step, showcasing how to handle three-dimensional integration.
Jacobian Determinant
The Jacobian determinant is crucial in coordinate transformation, such as moving from Cartesian to cylindrical coordinates. This determinant provides a scale factor that adjusts the size of the differential element when changing coordinates.

For the Cartesian to cylindrical transformation, we need the Jacobian since the volume element changes. In Cartesian coordinates, the differential volume element is given by \(dx\ dy\ dz\), which becomes \(r\ dr\ d\theta\ dz\) in cylindrical coordinates.
  • The transformation equations for x and y in cylindrical coordinates are \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
  • The Jacobian determinant evaluates to \(r\), which can be seen from simplifying the matrix determinant formed by partial derivatives.
Utilizing the Jacobian ensures the integration over this new coordinate system is accurate.
Coordinate Transformation
Coordinate transformation is a technique that simplifies integration by converting a complex integral into a more manageable form. By changing coordinates, we can take advantage of symmetry or other properties of the problem.

In this case, the transformation is from Cartesian coordinates \((x, y, z)\) to cylindrical coordinates \((r, \theta, z)\). This is particularly useful for regions that are easier to describe in terms of distance from an axis (the radius, r) and angle around that axis (the angle, \(\theta\)).

Cylindrical coordinates transform 3D problems involving cylinders or rotations around an axis into simpler calculations. This is done by expressing \(x\) and \(y\) through:
  • \(x = r\cos(\theta)\)
  • \(y = r\sin(\theta)\)
Transforming to cylindrical coordinates in our integral problem streamlined the process, as the boundaries of integration became constants, facilitating easier evaluation.
Multivariable Calculus
Multivariable calculus extends essential calculus concepts to functions of multiple variables. This branch deals with differentiation and integration of functions with more than one variable, crucial for exploration in several fields, including physics and engineering.

In this exercise, we deal with a triple integral, which is part of multivariable calculus. Triple integrals enable the calculation of volume-like quantities in three-dimensional spaces. They require integrating over three variables, which, in this case, correlate with radius, angle, and height (r, \(\theta\), z).
Performing these integrations requires careful consideration of the order and bounds for each integral. Here, the bounds of integration depend on geometric properties such as symmetry, embracing concepts of the region being integrated.
Understanding the bigger picture of how integration holds together under multiple variables, both conceptually and visually, greatly benefits solving intricate three-dimensional problems.

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Most popular questions from this chapter

Use a double integral to compute the area of the following regions. Make a sketch of the region. The region bounded by \(y=1+\sin x\) and \(y=1-\sin x\) on the interval \([0, \pi]\)

The Jacobian is a magnification (or reduction) factor that relates the area of a small region near the point \((u, v)\) to the area of the image of that region near the point \((x, y)\) a. Suppose \(S\) is a rectangle in the \(u v\) -plane with vertices \(O(0,0)\) \(P(\Delta u, 0),(\Delta u, \Delta v),\) and \(Q(0, \Delta v)\) (see figure). The image of \(S\) under the transformation \(x=g(u, v), y=h(u, v)\) is a region \(R\) in the \(x y\) -plane. Let \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) be the images of O, P, and \(Q,\) respectively, in the \(x y\) -plane, where \(O^{\prime}\) \(P^{\prime},\) and \(Q^{\prime}\) do not all lie on the same line. Explain why the coordinates of \(\boldsymbol{O}^{\prime}, \boldsymbol{P}^{\prime}\), and \(Q^{\prime}\) are \((g(0,0), h(0,0))\) \((g(\Delta u, 0), h(\Delta u, 0)),\) and \((g(0, \Delta v), h(0, \Delta v)),\) respectively. b. Use a Taylor series in both variables to show that $$\begin{aligned} &g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u\\\ &g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v\\\ &\begin{array}{l} h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array} \end{aligned}$$ where \(g_{u}(0,0)\) is \(\frac{\partial x}{\partial u}\) evaluated at \((0,0),\) with similar meanings for \(g_{v}, h_{u}\) and \(h_{v}\) c. Consider the vectors \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\) and the parallelogram, two of whose sides are \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\). Use the cross product to show that the area of the parallelogram is approximately \(|J(u, v)| \Delta u \Delta v\) d. Explain why the ratio of the area of \(R\) to the area of \(S\) is approximately \(|J(u, v)|\)

Two integrals to one Draw the regions of integration and write the following integrals as a single iterated integral: $$\int_{0}^{1} \int_{e^{y}}^{e} f(x, y) d x d y+\int_{-1}^{0} \int_{e^{-y}}^{e} f(x, y) d x d y$$

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and \(h\) are positive constants. Find the volume of a right circular cone with height \(h\) and base radius \(r\)

Let \(R\) be the region bounded by the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1,\) where \(a > 0\) and \(b>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v\). Find the center of mass of the upper half of \(R(y \geq 0)\) assuming it has a constant density.

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