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Chapter 14: Problem 22

Evaluate the following integrals as they are written. $$\int_{0}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} 2 x^{2} y d y d x$$

Short Answer

Expert verified
Question: Evaluate the double integral given by the integration of the function \(2x^2y\) with respect to \(y\) first and then with respect to \(x\). The limits of integration for \(y\) are given by \(-\sqrt{1-x^2}\) and \(\sqrt{1-x^2}\), and for \(x\) by \(0\) and \(1\). Answer: After evaluating the given double integral, the value is 0.

Step by step solution

01

Integrate with respect to y

First, we need to integrate the function \(2x^2y\) with respect to \(y\). As \(x^2\) is treated as a constant during this step, the integration becomes: $$\int 2x^2y dy = 2x^2 \int y dy = x^2y^2 + C_1$$
02

Apply limits of integration for y

Now, we will apply the given limits of integration for \(y\) which are \(-\sqrt{1-x^2}\) and \(\sqrt{1-x^2}\). We do this by finding the upper limit value minus the lower limit value: Upper limit value (when y = \(\sqrt{1-x^2}\)): $$x^2(\sqrt{1-x^2})^2 + C_1 = x^2(1-x^2) + C_1$$ Lower limit value (when y = \(-\sqrt{1-x^2}\)): $$x^2(-\sqrt{1-x^2})^2 + C_1 = x^2(1-x^2) + C_1$$ Applying the limits: $$x^2(1-x^2) + C_1 - (x^2(1-x^2) + C_1) = 0$$ As we can see, after applying the limits of integration for \(y\), the value of the integral becomes zero.
03

Integrate with respect to x

Now, we need to integrate the function \(0\) with respect to \(x\), essentially we will obtain: $$\int_0^1 0 dx = 0$$
04

Conclusion

After evaluating the given double integral, we find that the value is 0.

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Key Concepts

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

Integration Limits
Integration limits are essential for computing definite integrals, as they define the boundaries of the region over which integration occurs. In double integrals, we have two sets of limits: one for each variable.

In the given exercise, the outer integral represents the limits for the variable \(x\), ranging from 0 to 1. These limits enclose the span of \(x\)-values over which we evaluate the inner integral. The inner integral has limits for \(y\), which are expressed in terms of \(x\) as \(-\sqrt{1-x^{2}}\) to \(\sqrt{1-x^{2}}\). These limits describe a vertical slice through a circle with radius 1, centered at the origin.

  • Integration limits determine the starting and ending points for integration.
  • Inner limits may depend on the outer variable(s) to sculpt complex regions in the Cartesian plane.
  • Understanding the geometry of limits helps visualize the region over which you integrate.
Change of Variables
The change of variables, or variable substitution, is a technique often employed in integration. It involves substituting a different variable or expression for the original variable. This can make the problem simpler to solve and is particularly useful when integration limits or regions are complex.

In our exercise, while a change of variables is not directly employed, it is something to consider for intricate problems. Sometimes, describing limits in terms of new variables like polar or spherical coordinates can simplify the integration process:
  • Transforming variables can convert irregular regions into regular shapes, such as circles or ellipses, which are easier to work with.
  • New variables may lead to easier-to-integrate functions.
  • Always remember to update the differentials and compute the Jacobian if necessary during the transformation.
Iterated Integrals
An iterated integral is the process of evaluating double or multiple integrals by breaking them down into a sequence of single integrals. We first integrate with respect to one variable while treating others as constants, then repeat the process as necessary. This approach is standard when dealing with complex regions or functions.

In the problem, we first integrate the function \(2x^2y\) with respect to \(y\), treating \(x^2\) as a constant. Having integrated with respect to \(y\), we then apply the limits for \(y\) and finally integrate the resulting zero function with respect to \(x\):
  • The inner integral is performed first, focusing exclusively on one variable at a time.
  • After determining the inner integral, we solve the outer integral over the remaining variable.
  • This systematic approach reduces multidimensional integration into manageable single-variable problems.

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Most popular questions from this chapter

The Jacobian is a magnification (or reduction) factor that relates the area of a small region near the point \((u, v)\) to the area of the image of that region near the point \((x, y)\) a. Suppose \(S\) is a rectangle in the \(u v\) -plane with vertices \(O(0,0)\) \(P(\Delta u, 0),(\Delta u, \Delta v),\) and \(Q(0, \Delta v)\) (see figure). The image of \(S\) under the transformation \(x=g(u, v), y=h(u, v)\) is a region \(R\) in the \(x y\) -plane. Let \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) be the images of O, P, and \(Q,\) respectively, in the \(x y\) -plane, where \(O^{\prime}\) \(P^{\prime},\) and \(Q^{\prime}\) do not all lie on the same line. Explain why the coordinates of \(\boldsymbol{O}^{\prime}, \boldsymbol{P}^{\prime}\), and \(Q^{\prime}\) are \((g(0,0), h(0,0))\) \((g(\Delta u, 0), h(\Delta u, 0)),\) and \((g(0, \Delta v), h(0, \Delta v)),\) respectively. b. Use a Taylor series in both variables to show that $$\begin{aligned} &g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u\\\ &g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v\\\ &\begin{array}{l} h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array} \end{aligned}$$ where \(g_{u}(0,0)\) is \(\frac{\partial x}{\partial u}\) evaluated at \((0,0),\) with similar meanings for \(g_{v}, h_{u}\) and \(h_{v}\) c. Consider the vectors \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\) and the parallelogram, two of whose sides are \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\). Use the cross product to show that the area of the parallelogram is approximately \(|J(u, v)| \Delta u \Delta v\) d. Explain why the ratio of the area of \(R\) to the area of \(S\) is approximately \(|J(u, v)|\)

Let \(D\) be the region bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a > 0, b > 0\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v, z=c w\). Find the volume of \(D\)

Find the coordinates of the center of mass of the following solids with variable density. The interior of the prism formed by \(z=x, x=1, y=4,\) and the coordinate planes with \(\rho(x, y, z)=2+y\)

Many improper double integrals may be handled using the techniques for improper integrals in one variable. For example, under suitable conditions on \(f\) $$\int_{a}^{\infty} \int_{g(x)}^{h(x)} f(x, y) d y d x=\lim _{b \rightarrow \infty} \int_{a}^{b} \int_{g(x)}^{h(x)} f(x, y) d y d x$$ Use or extend the one-variable methods for improper integrals to evaluate the following integrals. $$\int_{1}^{\infty} \int_{0}^{1 / x^{2}} \frac{2 y}{x} d y d x$$

Evaluate the following integrals. $$\iint_{R} y d A ; R=\\{(x, y): 0 \leq y \leq \sec x, 0 \leq x \leq \pi / 3\\}$$
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