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Chapter 14: Problem 21

Find the volume of the solid that lies below the surface \(z=f(x, y)\) and above the region in the \(x y\) -plane bounded by the given curves. $$z=x^{2}+y^{2} ; x=0, x=1, y=0, y=2$$

Short Answer

Expert verified
The volume is \(\frac{10}{3}.\)

Step by step solution

01

Understand the Problem

We have a surface described by the equation \(z = x^2 + y^2\) and a region in the \(xy\)-plane that is defined by the boundaries \(x=0\), \(x=1\), \(y=0\), and \(y=2\). We need to find the volume enclosed by this surface and the \(xy\)-region.
02

Set Up the Double Integral

To find the volume under the surface \(z = x^2 + y^2\) and above the specified region, we will set up a double integral: \(\int_{0}^{1} \int_{0}^{2} (x^2 + y^2)\, dy \, dx.\) Here, the outer integral corresponds to \(x\) from 0 to 1, and the inner integral corresponds to \(y\) from 0 to 2.
03

Integrate with Respect to y

Compute the inner integral by integrating \(x^2 + y^2\) with respect to \(y\): \(\int_{0}^{2} (x^2 + y^2) \, dy.\)This results in: \(x^2y + \frac{y^3}{3}\) evaluated from \(y = 0\) to \(y = 2\).
04

Evaluate the y-Integral

Substitute the limits into the integrated function: \(\left(x^2(2) + \frac{2^3}{3} \right) - \left(x^2(0) + \frac{0^3}{3} \right) = 2x^2 + \frac{8}{3}.\)
05

Integrate with Respect to x

Compute the outer integral, \(\int_{0}^{1} \left( 2x^2 + \frac{8}{3} \right) \, dx.\)This gives: \(\frac{2}{3}x^3 + \frac{8}{3}x\) evaluated from \(x = 0\) to \(x = 1\).
06

Evaluate the x-Integral

Substitute the limits into the integrated function:\(\left(\frac{2}{3}(1)^3 + \frac{8}{3}(1) \right) - \left(\frac{2}{3}(0)^3 + \frac{8}{3}(0) \right) = \frac{2}{3} + \frac{8}{3} = \frac{10}{3}.\)
07

Conclusion

The final result represents the volume of the solid: \(\frac{10}{3}.\)

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Key Concepts

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

Volume Under Surface
In mathematics, finding the volume under a surface is a common task that involves using integration to calculate how much space is enclosed between a surface and a region on a plane beneath it. In this exercise, the surface is defined by the equation \(z = x^2 + y^2\). The goal is to determine the three-dimensional volume between this surface and the rectangular region on the \(xy\)-plane, which is defined by specific boundaries.
  • Surface Equation: The surface, \(z\), is determined by the function \(f(x,y) = x^2 + y^2\).
  • Region Boundaries: These are the limits within which we are interested in calculating the volume: \(x\) ranges from 0 to 1, and \(y\) ranges from 0 to 2.
This involves setting up a double integral that sums up infinitesimal contributions of volume over the entire region.
Bounded Regions
Bounded regions in calculus refer to specific areas on a plane where variables are constrained within certain limits. For double integrals, the bounded region determines where the function will be evaluated.
  • x-boundaries: From \(x = 0\) to \(x = 1\).
  • y-boundaries: From \(y = 0\) to \(y = 2\).
This region forms a rectangular shape on the \(xy\)-plane, providing the space over which the volume under the surface will be calculated. Choosing the correct bounds is crucial for an accurate integral setup, as they define where your results will be applicable.
Multivariable Calculus
In multivariable calculus, we extend calculus concepts to functions of more than one variable. Here, instead of dealing with a well-defined line or curve, we engage with surfaces and the quantity of space they encompass.
  • Functions of Two Variables: The given function, \(f(x, y) = x^2 + y^2\), is a function of two variables, \(x\) and \(y\).
  • 3D Space: This moves our focus from curves in 2D to surfaces in 3D, providing a richer set of tools for analysis.
Through double integration, multivariable calculus aids in finding areas, volumes, and sets the foundations for more advanced operations like surface integrals.
Integration Techniques
Integration techniques are fundamental in dealing with double integrals involving multiple steps, such as integrating a function over a specified region.
  • Set Up the Integral: Begin with setting up a double integral: \(\int_{0}^{1} \int_{0}^{2} (x^2 + y^2)\, dy \, dx\).
  • Integrate Respect to \(y\): First, integrate the inner integral with respect to \(y\): \(\int_{0}^{2} (x^2 + y^2) \, dy\).
  • Integrate Respect to \(x\): Then, integrate the resulting expression with respect to \(x\).
  • Evaluating Definite Integrals: Each integration step is followed by applying the limits, yielding a specific numeric result.
Using these techniques systematically, we arrive at the volume value \(\frac{10}{3}\), illustrating the power of integration in solving real-world spatial problems.

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Most popular questions from this chapter

Let \(R\) be the first-quadrant region bounded by the circles \(x^{2}+y^{2}=2 x, x^{2}+y^{2}=6 x\) and the circles \(x^{2}+y^{2}=2 y\) \(x^{2}+y^{2}=8 y\), Use the transformation $$ u=\frac{2 x}{x^{2}+y^{2}}, \quad y=\frac{2 y}{x^{2}+y^{2}} $$ to evaluate the integral $$ \iint_{R} \frac{1}{\left(x^{2}+y^{2}\right)^{2}} d x d y $$

Find the volume bounded by the elliptic paraboloids \(z=\) \(2 x^{2}+y^{2}\) and \(z=12-x^{2}-2 y^{2} .\) Note that this solid projects onto a circular disk in the \(x y\) -plane.

Sketch the solid bounded by the graphs of the given equations. Then find its volume by triple integration. $$z=x^{2}, \quad y+z=4, \quad y=0, \quad z=0$$

Find the centroid of the plane region bounded by the given curves. Assume that the density is \(\delta \equiv 1\) for each region. $$x=0, x=4, y=0, y=6$$

Use a computer algebra system first to plot and then to approximate (with four-place accuracy) the area of the part of the given surface \(S\) that lies above the square in the \(x y\) -plane defined by: (a) \(-1 \leqq x \leqq 1,-1 \leqq y \leqq 1\) : (b) \(|x|+|y| \leqq 1\) An ellipsoid with semiaxes \(a, b\), and \(c\) is defined by the parametrization \(x=a \sin \phi \cos \theta . \quad y=b \sin \phi \sin \theta, \quad z=c \cos \phi\) \((0 \leqq \phi \leqq \pi, 0 \leqq \theta \leqq 2 \pi)\) in terms of the angular spherical coordinates \(\phi\) and \(\theta .\) Use a computer algebra system to approximate (to four-place accuracy) the area of the ellipsoid with \(a=4, b=3\), and \(c=2\).
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