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Chapter 15: Problem 27

Find the volume of the solid lying under the elliptic paraboloid \(x^{2} / 4+y^{2} / 9+z=1\) and above the rectangle \(R=[-1,1] \times[-2,2]\)

Short Answer

Expert verified
The volume is \(\frac{184}{27}\) cubic units.

Step by step solution

01

Understand the region of integration

The elliptic paraboloid given is \(x^2/4 + y^2/9 + z = 1\). Solving for \(z\), it simplifies to \(z = 1 - x^2/4 - y^2/9\). The region \(R\) over which we integrate is the rectangle \([-1,1] \times [-2,2]\). We will integrate \(z\) over this region to find the volume.
02

Set up the double integral

The volume \(V\) is given by the double integral of \(z = 1 - \frac{x^2}{4} - \frac{y^2}{9}\) over the region \(R\). Thus, \[ V = \int_{-1}^{1} \int_{-2}^{2} \left(1 - \frac{x^2}{4} - \frac{y^2}{9}\right) \, dy \, dx. \]
03

Integrate with respect to \(y\)

First, we integrate the function \(1 - \frac{x^2}{4} - \frac{y^2}{9}\) with respect to \(y\) from \(-2\) to \(2\):\[ \int_{-2}^{2} \left( 1 - \frac{x^2}{4} - \frac{y^2}{9} \right) \, dy = \left[ y - \frac{x^2}{4}y - \frac{y^3}{27} \right]_{-2}^{2}. \]
04

Evaluate the inner integral

Substitute the limits for \(y\): \[ \left[ y - \frac{x^2}{4} y - \frac{y^3}{27} \right]_{-2}^{2} = \left( 2 - \frac{x^2}{2} - \frac{8}{27} \right) - \left( -2 + \frac{x^2}{2} + \frac{8}{27} \right) = 4 - \frac{16}{27}. \]
05

Integrate with respect to \(x\)

Now integrate the result from Step 4 with respect to \(x\) over \([-1, 1]\): \[ \int_{-1}^{1} \left(4 - \frac{16}{27} \right) \, dx = \int_{-1}^{1} \left( \frac{92}{27} \right) \, dx = \frac{92}{27} [x]_{-1}^{1}. \]
06

Evaluate the outer integral

Substitute the limits for \(x\):\[ \frac{92}{27} [x]_{-1}^{1} = \frac{92}{27} (1 - (-1)) = \frac{92}{27} \times 2 = \frac{184}{27}. \]
07

Final calculation and conclusion

The volume of the solid under the elliptic paraboloid and above the rectangle is \(\frac{184}{27}\) cubic units.

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Key Concepts

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

Double Integration
Double integration is a key concept in multivariable calculus. It allows us to compute the volume under a surface by integrating a function of two variables over a given region. In the context of this exercise, we are analyzing an elliptic paraboloid. To find the volume of the solid beneath it, we perform a double integral over a specified rectangle.
  • The double integral is set up as \( \int_{a}^{b} \int_{c}^{d} f(x, y) \, dy \, dx \). Here, \( f(x, y) \) is the function representing the surface.
  • The limits of the integral correspond to the bounds of the region over which we integrate.
  • First, the integration is performed with respect to one variable, usually the inner integral, then with respect to the other variable, the outer integral.
This stepwise approach helps determine the cumulative volume under the surface across the entire region of interest. Breaking this process into two parts simplifies calculations significantly and is a standard approach in finding volumes under complex surfaces.
Volume Under Surface
Finding the volume under a surface is a common application of double integration in multivariable calculus. In essence, you're looking at how much "space" lies below a function and above a defined region.
  • To determine this volume, you need two components: the equation describing the surface and the region in the xy-plane you are interested in.
  • Integration takes into account every tiny piece of area under the surface and sums them up over the region to find the total volume.
  • In our case, the surface is described by the elliptic paraboloid equation \( z = 1 - \frac{x^2}{4} - \frac{y^2}{9} \).
The double integral set up in the exercise processes this surface function over the rectangle \([-1, 1] \times [-2, 2]\) to compute the precise volume under this curved surface.
Elliptic Paraboloid
An elliptic paraboloid is a 3D surface that resembles a parabola extending in two dimensions. It has unique mathematical and geometric properties making it interesting for studying integration and volume.
  • The equation \( x^2/4 + y^2/9 + z = 1 \) describes this paraboloid. Rewriting it as \( z = 1 - x^2/4 - y^2/9 \) reveals how its height changes based on x and y values.
  • Elliptic paraboloids curve upwards, and you can visualize them like satellite dishes opening towards the sky.
  • By integrating below such surfaces, we're effectively measuring how much space is enclosed by its curve over a certain plane region.
Understanding the parabolic nature of this surface helps in comprehending why certain integration techniques, such as changing bounds, may or may not apply based on symmetry and curvature.
Integral Calculus
Integral calculus is a fundamental branch of mathematics. It focuses on finding the total size or value of properties bound within a space. This is achieved by summing infinitesimally small pieces.
  • The concept of integration extends to functions of multiple variables, offering insights into areas, volumes, and other higher-dimensional properties.
  • In double integrals, each integration accounts for one of two variables in a planar region, allowing us to gain insights into three-dimensional aspects like volume.
  • Practicing integration with functions, such as elliptic paraboloids, enhances understanding of how integral calculus captures real-world quantity measurements.
By using integral calculus, particularly double integration, students unpack the complex interaction of multiple variables and begin to visualize abstract mathematical concepts into tangible three-dimensional analyses.

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Most popular questions from this chapter

(a) Evaluate \(\iiint_{E} d V,\) where \(E\) is the solid enclosed by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1 .\) Use the transformation \(x=a u, y=b v, z=c w\) (b) The earth is not a perfect sphere; rotation has resulted in flattening at the poles. So the shape can be approximated by an ellipsoid with \(a=b=6378 \mathrm{km}\) and \(c=6356 \mathrm{km} .\) Use part (a) to estimate the volume of the earth.

Use the given transformation to evaluate the integral. \(\iint_{R} x y d A,\) where \(R\) is the region in the first quadrant bounded by the lines \(y=x\) and \(y=3 x\) and the hyperbolas \(x y=1\) \(x y=3 ; \quad x=u / v, y=v\)

Use a computer algebra system to find the exact volume of the solid. Enclosed by \(z=1-x^{2}-y^{2}\) and \(z=0\)

Use the Midpoint Rule for triple integrals (Exercise 24 to estimate the value of the integral. Divide \(B\) into eight sub-boxes of equal size. $$ \begin{array}{l}{\int \mathbb{N}_{B} \frac{1}{\ln (1+x+y+z)} d V, \text { where }} \\ {B=\\{(x, y, z) | 0 \leqslant x \leqslant 4,0 \leqslant y \leqslant 8,0 \leqslant z \leqslant 4\\}}\end{array} $$

Sketch the solid whose volume is given by the iterated integral. $$ \int_{0}^{2} \int_{0}^{2-y} \int_{0}^{4-y^{2}} d x d z d y $$
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