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Chapter 14: Problem 15

Find all values of the constant \(c\) so that \(\int_{0}^{1} \int_{0}^{c}(2 x+y) d x d y=3\).

Short Answer

Expert verified
The value of \(c\) is \(\frac{3}{2}\).

Step by step solution

01

Set up the double integral

Begin by setting up the double integral given in the problem statement: \[\int_{0}^{1} \int_{0}^{c}(2x+y) \, dx \, dy = 3\] This is our integral that needs to add up to 3.
02

Evaluate the inner integral

Evaluate the inner integral that involves integrating with respect to \(x\):\[\int_{0}^{c}(2x+y) \, dx\]This integral evaluates to:\[\left[ x^2 + xy \right]_{0}^{c} = (c^2 + cy) - (0+0) = c^2 + cy\]
03

Set up and evaluate the outer integral

Insert the result from the inner integral into the outer integral:\[\int_{0}^{1} (c^2 + cy) \, dy\]Evaluate this integral:\[\left[ c^2y + \frac{c}{2}y^2 \right]_{0}^{1} = c^2(1) + \frac{c}{2}(1)^2 - (c^2(0) + \frac{c}{2}(0)^2)\]This simplifies to:\[c^2 + \frac{c}{2}\]
04

Set the result equal to 3 and solve for c

We know from the problem statement that this value equals 3:\[c^2 + \frac{c}{2} = 3\]Multiply through by 2 to clear the fraction:\[2c^2 + c = 6\]Rearrange it to form a quadratic equation:\[2c^2 + c - 6 = 0\]
05

Solve the quadratic equation

Use the quadratic formula to solve for \(c\):\[c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 2\), \(b = 1\), \(c = -6\). Substituting these values gives:\[c = \frac{-1 \pm \sqrt{1 + 48}}{4}\]This simplifies to:\[c = \frac{-1 \pm \sqrt{49}}{4}\]Thus, \(\sqrt{49} = 7\), so:\[c = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2}\] or:\[c = \frac{-1 - 7}{4} = -2\]
06

Evaluate the solutions

Check the context of the problem. Since \(c\) represents the upper limit of integration, a negative value would not make sense in this context. Thus, only the positive solution is relevant.

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Key Concepts

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

Definite Integration
Definite integration involves finding the value of an integral with specific bounds. These "bounds" or "limits of integration" specify the start and end points over which you are integrating.
In the double integral provided in the problem, these bounds are given for both the x and y variables:
  • The inner integral has the limit from 0 to \(c\).
  • The outer integral ranges from 0 to 1.
The goal of definite integration is to calculate the total "accumulation" of a function over the specified interval. This is particularly useful for calculating areas under curves. In the exercise, you integrated the function \((2x + y)\) to find a specific total, which was 3.
Unlike indefinite integrals that include a variable constant \(C\), definite integrals result in a specific number because the areas are well-defined by their limits.
Quadratic Equations
Quadratic equations play a crucial role when evaluating many integrals, including the exercise provided. A quadratic equation is generally expressed in the form: \(ax^2 + bx + c = 0\).
In this exercise, the solution led to the formation of a quadratic equation:
  • \(2c^2 + c - 6 = 0\)
Solving this quadratic equation involved using the quadratic formula, a tool used for finding the roots (or solutions) of the equation. The quadratic formula is defined as:\[c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, parameters \(a\), \(b\), and \(c\) correspond to the coefficients in the quadratic equation. In this context:
  • \(a = 2\)
  • \(b = 1\)
  • \(c = -6\)
The formula provides two potential solutions, known as roots. In mathematics, quadratic equations often have two solutions, but only relevant solutions should be considered based on the context of the problem.
For this integral, only the positive root \(\frac{3}{2}\) was suitable, as \(c\) related to a dimensional property (the upper bound of the inner integral).
Limits of Integration
Limits of integration are essential in specifying the region over which you perform integration. In multiple integrals, knowing these bounds is crucial for the proper calculation of areas or volumes.
In the given exercise, the integral's limits are clearly defined:
  • For the inner integral, \(x\) varies from 0 to \(c\).
  • For the outer integral, \(y\) spans from 0 to 1.
These bounds delineate a region on the xy-plane over which the function \(2x + y\) is summed. Limits help in determining how much of this function contributes to the total, represented by the number 3 in this problem.
In context, limits of integration are essential in ensuring the resulting integral reflects the desired geometric or functional characteristics. Proper setup and understanding of these limits are vital. They ensure the equation models the problem correctly, particularly with double integrals involving nested bounds, such as with \((x, y)\) in this problem.

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Most popular questions from this chapter

Inverse transform The equations \(x=g(u, v), y=h(u, v)\) in Figure 14.56 transform the region \(G\) in the \(u v\) -plane into the region \(R\) in the \(x y\) -plane. since the substitution transformation is one-to-one with continuous first partial derivatives, it has an inverse transformation, and there are equations \(u=\alpha(x, y), v=\beta(x, y)\) with continuous first partial derivatives transforming \(R\) back into \(G .\) Moreover, the Jacobian determinants of the transformations are related reciprocally by $$\frac{\partial(x, y)}{\partial(u, v)}=\left(\frac{\partial(u, v)}{\partial(x, y)}\right)^{-1}$$ Equation (10) is proved in advanced calculus. Use it to find the area of the region \(R\) in the first quadrant of the \(x y\) -plane bounded by the lines \(y=2 x, 2 y=x,\) and the curves \(x y=2,2 x y=1\) for \(u=x y\) and \(v=y / x.\)

Let \(D\) be the region in the first octant that is bounded below by the cone \(\phi=\pi / 4\) and above by the sphere \(\rho=3 .\) Express the volume of \(D\) as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find \(V\)

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.7 .\) Evaluate the improper integrals as iterated integrals. $$\int_{0}^{\infty} \int_{0}^{\infty} x e^{-(x+2 y)} d x d y$$

Find the volume of the region cut from the solid cylinder \(x^{2}+y^{2} \leq 1\) by the sphere \(x^{2}+y^{2}+z^{2}=4\)

Find the average value of the function \(f(\rho, \phi, \theta)=\rho \cos \phi\) over the solid upper ball \(\rho \leq 1,0 \leq \phi \leq \pi / 2\)
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