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Chapter 15: Problem 30

Find the volume of the solid in the first octant bounded by the cylinder \(z=16-x^{2}\) and the plane \(y=5 .\)

Short Answer

Expert verified
Volume is \(\frac{640}{3}\) cubic units.

Step by step solution

01

Identify the Region

We are dealing with a 3D region in the first octant that is bounded by a cylinder and a plane. The cylinder is represented by the equation \(z = 16 - x^2\), indicating that it is an upward opening paraboloid with vertex at \((0, 0, 16)\). The plane \(y = 5\) is vertical and parallel to the \(xz\)-plane.
02

Determine the Bounds for x and y

Since we are in the first octant, \(x\) and \(y\) are non-negative. The plane \(y = 5\) bounds \(y\) from above, so \(0 \leq y \leq 5\). The paraboloid intersects the \(xz\)-plane when \(z = 0\), giving us \(x^2 = 16\), thus the bounds for \(x\) are \(-4 \leq x \leq 4\), but since we are in the first octant, \(0 \leq x \leq 4\).
03

Establish the Integral for Volume

The volume of the solid can be found by integrating the function \(z = 16 - x^2\) over the appropriate region. Since \(z\) varies from \(0\) to \(16 - x^2\), the integral for calculating volume is given by: \[ V = \int_{0}^{5} \int_{0}^{4} (16 - x^2) \, dx \, dy \].
04

Integrate with Respect to x

First, evaluate the inner integral with respect to \(x\):\[ \int_{0}^{4} (16 - x^2) \, dx = \left[ 16x - \frac{x^3}{3} \right]_{0}^{4} = (16 \times 4 - \frac{4^3}{3}) = 64 - \frac{64}{3} = \frac{128}{3} \].
05

Integrate with Respect to y

Now, integrate the result with respect to \(y\):\[ \int_{0}^{5} \frac{128}{3} \, dy = \frac{128}{3}y \bigg|_{0}^{5} = \frac{128}{3} \times 5 = \frac{640}{3} \].
06

Conclusion and Interpretation of Results

The calculated volume of the solid bounded by the cylinder and the plane in the first octant is \(\frac{640}{3}\). This represents the space enclosed within those constraints.

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Key Concepts

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

Volume of Solid
Finding the volume of a solid in a three-dimensional space is an important application of calculus. In this context, the solid is formed by a combination of surfaces: a cylinder and a plane. The term "solid" refers to the space bounded by these surfaces.
To calculate the volume of such a solid, we set up an integral that spans the three spatial dimensions, covering the entire region confined by those bounds.
In our specific problem, the volume is limited to the first octant, meaning only non-negative values for the coordinate axes are considered. This simplifies the calculation since it reduces the feasible region for the integration, ensuring we only consider the first part of the space where the cylinder and plane enclose a finite volume.
This volume is then determined by setting up integrals that accumulate values over the designated space, which leads us to the concept of triple integration, helping to find the total volume within the bounds.
Triple Integration
Triple integration is a method in calculus used to calculate volumes of 3D regions, among other applications. It extends the concept of double integration (used for finding areas in 2D) to three dimensions by integrating a function over a region in 3D space.
When dealing with triple integrals, you evaluate iterated integrals. This process involves integrating a nested series of integrals, each corresponding to one variable of the function in the 3D region of interest.
In this exercise, we simplify the 3D volume calculation by first integrating with respect to the innermost variable, which is often chosen based on the function's limits鈥攈ere, this variable is x. Once this inner integral has been solved, the result is integrated with respect to the next variable, y, completing the volume computation by covering all directions over the defined boundaries.
  • The first integration, with respect to one variable while treating others as constants, reduces part of the volume to a 2D problem.
  • The subsequent integrations account for the remaining dimensions, accumulating the volume incrementally.
The approach used here highlights how we decompose complex geometry into manageable pieces, applying integration iteratively to capture the essence of volume in a multi-variable scenario.
Integration Bounds
Integration bounds are crucial since they define the "limits" of integration. They specify over what region of space the integral should be calculated. These bounds ensure that calculations remain within the specified space, which is essential when calculating properties like volume.
In our problem, the bounds are clearly outlined by:
  • A cylinder defined by the equation \( z = 16 - x^2 \), which limits the height of the solid in the direction of the z-axis.
  • A plane \( y = 5 \), which provides the maximum extent of the solid in the y-direction.
  • Being restricted to the first octant indicates that both \( x \) and \( y \) are non-negative, meaning \( x \geq 0 \) and \( y \geq 0 \).
To integrate over the region, these bounds are translated into the limits of the integrals, first setting \( 0 \leq x \leq 4 \) based on the paraboloid's domain and then setting \( 0 \leq y \leq 5 \). Each integration process adheres to its specified limits, ensuring the correct finite volume within these bounds is calculated, capturing the specified segment of space.

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Most popular questions from this chapter

\(21-34\) Use spherical coordinates. Evaluate \(\iiint_{E} x y z d V,\) where \(E\) lies between the spheres \(\rho=2\) and \(\rho=4\) and above the cone \(\phi=\pi / 3\)

(a) A lamp has two bulbs of a type with an average lifetime of 1000 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean \(\mu=1000,\) find the probontity that both of the lamp's bulbs fail within 1000 hours. (b) Another lamp has just one bulb of the same type as in part (a). If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1000 hours.

(a) We define the improper integral (over the entire plane \(\mathbb{R}^{2}\) ) $$\begin{aligned} I &=\iint e^{-\left(x^{2}+y^{2}\right)} d A=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d y d x \\ &=\lim _{a \rightarrow \infty} \iint e^{-\left(x^{2}+y^{2}\right)} d A \end{aligned}$$ where \(D_{a}\) is the disk with radius \(a\) and center the origin. Show that $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d A=\pi$$ (b) An equivalent definition of the improper integral in part (a) is $$\iint_{\mathbb{R}^{\prime}} e^{-\left(x^{2}+y^{2}\right)} d A=\lim _{a \rightarrow \infty} \iint_{S_{a}} e^{-\left(x^{2}+y^{2}\right)} d A$$ where \(S_{,}\) is the square with vertices \((\pm a, \pm a) .\) Use this to show that $$int_{-\infty}^{\infty} e^{-x^{2}} d x \int_{-\infty}^{\infty} e^{-y^{2}} d y=\pi$$ (c) Deduce that $$\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$$ (d) By making the change of variable \(t=\sqrt{2} x,\) show that $$\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}$$ (This is a fundamental result for probability and statistics.)

Use cylindrical coordinates. (a) Find the volume of the region \(E\) bounded by the paraboloids \(z=x^{2}+y^{2}\) and \(z=36-3 x^{2}-3 y^{2}\) . (b) Find the centroid of \(E\) (the center of mass in the case where the density is constant).

The latitude and longitude of a point \(P\) in the Northern Hemisphere are related to spherical coordinates \(\rho, \theta, \phi\) as follows. We take the origin to be the center of the earth and the positive \(z\) -axis to pass through the North Pole. The positive \(x\) -axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of \(P\) is \(\alpha=90^{\circ}-\phi^{\circ}\) and the longitude is \(\beta=360^{\circ}-\theta^{\circ} .\) Find the great-circle distance from Los Angeles (lat. \(34.06^{\circ} \mathrm{N},\) long. \(118.25^{\circ} \mathrm{W}\) ) to Montreal (lat. \(45.50^{\circ} \mathrm{N},\) long. \(73.60^{\circ} \mathrm{W} ) .\) Take the radius of the earth to be 3960 \(\mathrm{mi} .\) (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.)
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