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Chapter 13: Problem 22

Find the volume of the following solids using triple integrals. The solid bounded by the surfaces \(z=e^{y}\) and \(z=1\) over the rectangle \(\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq \ln 2\\}\)

Short Answer

Expert verified
Answer: The volume of the solid is \(V = 3 - \ln 2\).

Step by step solution

01

Determine the order of integration and the limits of integration

We need to set up the integral in terms of \(x\), \(y\), and \(z\). Since the rectangle is given in terms of \(x\) and \(y\), it's natural to choose the order of integration as \(dx\, dy\, dz\). The limits of integration for \(x\) and \(y\) are given by the rectangle: \(0 \leq x \leq 1\) and \(0 \leq y \leq \ln 2\). For the \(z\) limits, we must determine the range of \(z\) values for the solid. The solid is bounded by the surfaces \(z=e^{y}\) and \(z=1\), so the \(z\) limits are \(1 \leq z \leq e^{y}\).
02

Set up the triple integral

Now we will set up the triple integral to find the volume of the solid: $$ V = \int\int\int dx\,dy\,dz $$ Using the limits of integration from Step 1, we get: $$ V = \int_{0}^{\ln 2}\int_{0}^{1}\int_{1}^{e^y} dx\,dy\,dz $$
03

Evaluate the triple integral

Now we will evaluate the triple integral one variable at a time, starting with \(z\): $$ V = \int_{0}^{\ln 2}\int_{0}^{1}\left[\int_{1}^{e^y} dz\right] dx\,dy $$ $$ V = \int_{0}^{\ln 2}\int_{0}^{1} (e^y - 1) dx\, dy $$ Now we will evaluate the integral with respect to \(x\): $$ V = \int_{0}^{\ln 2}\left[\int_{0}^{1} (e^y - 1) dx\right] dy $$ $$ V = \int_{0}^{\ln 2} (e^y - 1) dy $$ Finally, we will evaluate the integral with respect to \(y\): $$ V = \left[e^y - y\right]_{0}^{\ln 2} $$ $$ V = (4 - \ln 2) - (1-0) $$ $$ V = 3 - \ln 2 $$ So, the volume of the solid bounded by the surfaces \(z=e^{y}\) and \(z=1\) over the rectangle \(\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq \ln 2\\}\) is \(V = 3 - \ln 2\).

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Key Concepts

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

Volume Calculation Using Triple Integrals
Volume calculation with triple integrals represents a fundamental aspect of multivariable calculus. It enables us to find the volume of a three-dimensional region by dividing it into infinitesimally small elements and summing their contributions. Imagine slicing a loaf of bread into very thin pieces: with triple integrals, we mathematically slice the space into tiny volumes, quantify each, and sum them all.

Think of the volume element, often written as dV or in this case as dx\text{ }dy\text{ }dz, which represents a small cube with dimensions dx by dy by dz. By integrating this tiny volume across the entire region of interest, we calculate the total volume enclosed by the given boundaries. In our textbook example, the integral V = \[ \int \int \int dx\text{ }dy\text{ }dz \] is set up to encompass the solid region bounded by specific surfaces and over a certain rectangle in the xy-plane.
Determining Integration Limits
In order to successfully use triple integrals for volume calculation, one must correctly determine the integration limits for each variable. These limits define the bounds within which we are integrating, similar to placing fences around the area where you're counting apples in an orchard.

To set the integration limits:
  • Look at the given bounding surfaces and shapes in the problem. For example, our solid is limited by z=e^y and z=1 in one direction and by a rectangle in the xy-plane.
  • Determine the range of each variable, often guided by the geometry or constraints provided in the problem. In our case, x varies from 0 to 1, y varies from 0 to ln 2, and z has a range between 1 and e^y as dictated by the bounding surfaces.
  • Finally, decide on the order of integration, which can be influenced by the simplicity of the limits or the integrand; commonly, it follows a dz dy dx or dx dy dz pattern, but can be adapted as needed.
A correct assessment of these limits is crucial for setting up the integral that will lead to the desired volume.
Evaluation of Triple Integrals
Evaluating triple integrals is a systematic process that requires us to work one variable at a time, from the innermost integral to the outermost, each time performing the integration over the interval set by our integration limits.

Integrating Step-By-Step

In the exercise, we start with z, then move to x, and finally integrate with respect to y. Each step simplifies the function gradually until we're left with a function of a single variable. This last integral, once evaluated, gives the final volume.

Common pitfalls to avoid when evaluating triple integrals include:
  • Mistaking the order of integration, which could make the problem more complicated or even unsolvable.
  • Forgetting to update the limits of integration when the order of integration is swapped.
  • Not evaluating step by step and attempting to integrate over all variables at once, which could lead to errors.
  • Misinterpreting the physical region of integration, leading to incorrect limits and therefore incorrect results.
Keeping these steps in mind and practicing meticulous evaluation of each integral will lead to a successful volume calculation, as shown in the solution with a final result of V = 3 - ln 2 for our specific problem.

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Most popular questions from this chapter

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\) \((\bar{x}, \bar{y})=\left(\frac{128}{105 \pi}, \frac{128}{105 \pi}\right)$$(\bar{x}, \bar{y})=\left(\frac{17}{18}, 0\right)\)

Changing order of integration If possible, write iterated integrals in cylindrical coordinates for the following regions in the specified orders. Sketch the region of integration. The solid above the cone \(z=r\) and below the sphere \(\rho=2,\) for \(z \geq 0,\) in the orders \(d z d r d \theta, d r d z d \theta,\) and \(d \theta d z d r\)

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid is enclosed by a hemisphere of radius \(a\). How far from the base is the center of mass?

Meaning of the Jacobian 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 $$\begin{array}{l} g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u \\ g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v \\ h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array}$$ 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 \(|J(u, v)|\)

Limiting center of mass \(A\) thin rod of length \(L\) has a linear density given by \(\rho(x)=\frac{10}{1+x^{2}}\) on the interval \(0 \leq x \leq L\). Find the mass and center of mass of the rod. How does the center of mass change as \(L \rightarrow \infty ?\)
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