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Chapter 14: Problem 29

Use a double integral to find the volume. The volume under the plane \(z=2 x+y\) and over the rectangle \(R=\\{(x, y): 3 \leq x \leq 5,1 \leq y \leq 2\\}\)

Short Answer

Expert verified
The volume under the plane is 19 cubic units.

Step by step solution

01

Identify the function and the region

The function given is the plane described by the equation \(z=2x+y\). The region of integration is defined as a rectangle \(R\) in the \(xy\)-plane, with corners at \((3,1)\), \((3,2)\), \((5,1)\), and \((5,2)\). This region can be written as \(3 \leq x \leq 5\) and \(1 \leq y \leq 2\).
02

Set up the double integral

To find the volume under the plane \(z=2x+y\) over the region \(R\), we set up the double integral as follows: \[V = \int_{1}^{2} \int_{3}^{5} (2x + y) \, dx \, dy\] This integral will give us the volume between the surface \(z = 2x + y\) and the rectangle on the \(xy\)-plane.
03

Integrate with respect to x

First, integrate the function \(2x+y\) with respect to \(x\), treating \(y\) as a constant:\[\int_{3}^{5} (2x + y) \, dx = \left[ x^2 + yx \right]_{3}^{5} = ((5^2 + 5y) - (3^2 + 3y)) = (25 + 5y) - (9 + 3y) = 16 + 2y\]
04

Integrate with respect to y

Now integrate the result \(16 + 2y\) with respect to \(y\):\[\int_{1}^{2} (16 + 2y) \, dy = \left[ 16y + y^2 \right]_{1}^{2} = ((16 \cdot 2 + 2^2) - (16 \cdot 1 + 1^2)) = (32 + 4) - (16 + 1) = 36 - 17 = 19\]
05

Conclusion

The volume of the solid under the plane \(z=2x+y\) over the rectangle \(R\) is given by the value of the double integral. We've found that this volume is 19 units cubed.

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Key Concepts

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

Volume Calculation
When finding the volume of a solid defined under a surface, like in this exercise, double integration is often used. Here, the surface is defined by the plane equation \( z = 2x + y \), and the area of interest is the rectangle \( R \) with boundaries in the \( xy \)-plane. To calculate the volume under this plane over the rectangle, we use a double integral.The double integral serves as a tool to accumulate all the infinite infinitely small volume elements under the surface over the region \( R \). Each small volume is essentially a thin slice or slab with height equal to the value of the surface at that point and a base area defined by an infinitesimal rectangle \( dx \, dy \). When summed over the area, these create the overall volume. For the given problem, this led us to calculate:\[ V = \int_{1}^{2} \int_{3}^{5} (2x + y) \, dx \, dy \].
Integration Techniques
Integration techniques become pivotal in determining the resultant value from a double integral. Here, we follow a two-step process:
  • First, integrate with respect to \( x \) while treating \( y \) as a constant. This approach is known as iterated integration, and it simplifies the problem into one-dimensional slices one variable at a time. The result of this integral forms a function of \( y \).
  • Next, integrate this \( y \)-dependent result over \( y \). This fully evaluates the integral over the designated region, providing the total volume.
In our exercise, integrating \( 2x + y \, dx \) from \( x=3 \) to \( x=5 \) yielded \( 16 + 2y \). Integrating \( 16 + 2y \) from \( y=1 \) to \( y=2 \) determined the final volume, resulting in a straightforward arithmetic solution: the volume is 19 cubic units.
Multivariable Calculus
Multivariable calculus extends the concepts of differentiation and integration to functions of multiple variables. In the context of this exercise, it allows us to handle the function \( z = 2x + y \) depending on two variables: \( x \) and \( y \). This function provides the height of the surface at any point within the rectangular region in the \( xy \)-plane.Applying double integrals is a common method in multivariable calculus when the intention is to calculate the volume under a surface or across a region in a plane. The process involves integrating a multivariable function across bounded domains such as rectangles or more irregular shapes.Understanding the relationship between volume and multivariable functions not only involves tackling such problems with integration techniques but also enhances understanding of geometric interpretations, enabling us to visualize how functions behave across different dimensions.

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Most popular questions from this chapter

Evaluate the iterated integral. $$ \int_{0}^{\pi / 2} \int_{0}^{\cos \theta} r^{3} d r d \theta $$

These exercises reference the Theorem of Pappus: If \(R\) is a bounded plane region and \(L\) is a line that lies in the plane of \(R\) such that \(R\) is entirely on one side of \(L,\) then the volume of the solid formed by revolving \(R\) about \(L\) is given by $$\text {volume}=(\text {area of } R) \cdot\left(\begin{array}{c}{\text {distance} \text {traveled}} \\ {\text {by the centroid}}\end{array}\right)$$ Use the Theorem of Pappus to find the centroid of the triangular region with vertices \((0,0),(a, 0),\) and \((0, b),\) where \(a>0\) and \(b>0 .[\text {Hint}: \text { Revolve the region about the } x-\) axis to obtain \(\bar{y} \text { and about the } y \text { -axis to obtain } \bar{x} .]\)

Use the transformation \(u=y / x, v=x y\) to find $$ \iint_{R} x y^{3} d A $$ over the region \(R\) in the first quadrant enclosed by \(y=x\) \(y=3 x, x y=1, x y=4\)

Find the center of gravity of the cube that is determined by the inequalities \(0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\) if (a) the density is proportional to the square of the distance to the origin; (b) the density is proportional to the sum of the distances to the faces that lie in the coordinate planes.

The transformation \(x=a u, y=b v(a>0, b>0)\) can be rewritten as \(x / a=u, y / b=v,\) and hence it maps the circular region $$ u^{2}+v^{2} \leq 1 $$ into the elliptical region $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 $$ In these exercises, perform the integration by transforming the elliptical region of integration into a circular region of integration and then evaluating the transformed integral in polar coordinates. \(\iint_{R} e^{-\left(x^{2}+4 y^{2}\right)} d A,\) where \(R\) is the region enclosed by the ellipse \(\left(x^{2} / 4\right)+y^{2}=1\)
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