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Chapter 9: Problem 35

\(\int_{0}^{1} \int_{x}^{1} x^{2} \sqrt{1+y^{4}} d y d x=\int_{0}^{1} \int_{0}^{y} x^{2} \sqrt{1+y^{4}} d x d y=\left.\int_{0}^{1} \frac{1}{3} x^{3} \sqrt{1+y^{4}}\right|_{0} ^{y} d y\) \(=\frac{1}{3} \int_{0}^{1} y^{3} \sqrt{1+y^{4}} d y=\left.\frac{1}{3}\left[\frac{1}{6}\left(1+y^{4}\right)^{3 / 2}\right]\right|_{0} ^{1}=\frac{1}{18}(2 \sqrt{2}-1)\)

Short Answer

Expert verified
The value of the integral is \(\frac{1}{18} (2\sqrt{2} - 1)\).

Step by step solution

01

Understand the Double Integral

The given problem involves evaluating a double integral:\[\int_{0}^{1} \int_{x}^{1} x^{2} \sqrt{1+y^{4}} d y d x\] This integral represents the volume under the surface \(x^2 \sqrt{1+y^4}\) over a specific region. Our goal is to evaluate this double integral by changing the order of integration.
02

Convert the Integral Limits

We start by considering the region of integration, which is bounded by \(x\) and \(1\) for \(y\), and \(0\) to \(1\) for \(x\). By reversing the order of integration, the integral is transformed to:\[\int_{0}^{1} \int_{0}^{y} x^{2} \sqrt{1+y^{4}} d x d y\]This reverses the integration with respect to \(x\) and \(y\), requiring us to first integrate with respect to \(x\), then \(y\).
03

Integrate with Respect to x

With the new limits, the inner integral is:\[\int_{0}^{y} x^{2} \sqrt{1+y^{4}} d x\]Here, \(\sqrt{1+y^4}\) is constant with respect to \(x\), so the integration is straightforward:\[\sqrt{1+y^4} \int_{0}^{y} x^{2} d x\]Using the power rule, we find the antiderivative of \(x^2\), which gives \(\frac{1}{3}x^3\). Evaluating from \(0\) to \(y\) gives:\[\frac{1}{3} y^3 \sqrt{1+y^4}\]
04

Integrate with Respect to y

Now we integrate the expression we just found:\[\frac{1}{3} \int_{0}^{1} y^{3} \sqrt{1+y^{4}} d y\]We continue by evaluating this new single integral, which involves a non-standard form. The guide suggests using a transformation or lookup method, known to evaluate such forms.
05

Evaluate the Final Integral

The provided solution involves substituting and finding antiderivatives through evaluated transforms or known results:\[\frac{1}{3} \left.\left[ \frac{1}{6} \left(1+y^{4}\right)^{3/2} \right] \right|_{0}^{1}\]Calculating this step-by-step, when \(y = 1\):\[\frac{1}{6} (1+1^4)^{3/2} = \frac{1}{6} (2)^{3/2} = \frac{\sqrt{2}}{3}\]When \(y = 0\), the expression simplifies to zero, so we have:\[\frac{\sqrt{2}}{3} \times \frac{1}{3} = \frac{1}{18} (2\sqrt{2} - 1)\]
06

Conclude with the Final Answer

Combining all the evaluated steps, the solution results in:\[\frac{1}{18} (2\sqrt{2} - 1)\]This represents the value of the double integral, taking into account both regions of integration and their conversions.

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Key Concepts

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

Order of Integration
Understanding the order of integration in double integrals is crucial for evaluating them successfully. In the original problem, the given double integral is presented with a specific order:
  • The outer integral is with respect to variable \(x\).
  • The inner integral is with respect to variable \(y\).
Switching the order of integration often simplifies the process of solving the integral. By converting the integral limits, we change the order to have:
  • The outer integral is with respect to variable \(y\).
  • The inner one with respect to \(x\).
This switch, which involves reversing the integration bounds and interchanging \(x\) and \(y\), can be particularly helpful when one set of integral limits is easier to evaluate or when getting rid of complex placeholder functions. Mastering this process requires understanding how to interpret and change regions of integration geometrically.
Volume Under Surface
In the context of double integrals, a common application is determining the volume under a surface. Consider the function from the original problem: \(x^2 \sqrt{1+y^4}\). This function defines a surface in three-dimensional space. The double integral of this function over a designated region calculates the volume of the prism, or the solid that lies beneath this surface and above a specified area in the \(xy\)-plane. This is achieved by accumulating small volumes under the surface across infinitesimal rectangles in this area. By integrating, we sum up these small volumes:
  • The inner integral calculates the volume of these slices in one direction.
  • The outer integral sums these slices over the entire specified region, completing the volume calculation.
This process beautifully demonstrates the power and practical application of double integrals in geometrical problem solving.
Antiderivative
The concept of an antiderivative is pivotal in the process of solving integrals. It is essentially the reverse process of derivation. To deal with an integral, finding the antiderivatives of the involved functions is crucial.In the third step of solving the original problem, the inner integral \(\int_{0}^{y} x^{2} dx\) requires us to find the antiderivative of \(x^2\). Here, applying the power rule yields:
  • \(\frac{1}{3} x^3\)
Evaluating this from the bounds of 0 to \(y\) simple mechanical calculation results in \(\frac{1}{3} y^3\).Antiderivatives make it possible to evaluate definite integrals, thereby converting expressions in standard forms into useful values. Mastery of antiderivatives is essential for anyone dealing with calculus, as they are crucial for both single and multiple integrals.
Non-Standard Integral Forms
Handling non-standard integral forms can sometimes be challenging and may require special techniques or transformations. These integrals often do not align with the typical integral forms that are easily solvable.In the final step of the original problem, the integral \(\frac{1}{3} \int_{0}^{1} y^{3} \sqrt{1+y^{4}} dy\) involves such a form. Solving it analytically requires leveraging substitutions or known results.Specific techniques like trigonometric substitution or using tabulated integral solutions can be significant aids.
  • In cases with complex forms like \(\sqrt{1+y^4}\), looking up tables or applying a transformation, such as a change of variables, simplifies the integration.
Cultivating the ability to navigate non-standard forms opens up new realms of problem-solving possibilities, especially in advanced calculus.

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Most popular questions from this chapter

\(\operatorname{div} \mathbf{F}=2 x+2 z+12 z^{2}\) $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} \operatorname{div} \mathbf{F} d V=\int_{0}^{3} \int_{0}^{2} \int_{0}^{1}\left(2 x+2 z+12 z^{2}\right) d x d y d z \\\&=\left.\int_{0}^{3} \int_{0}^{2}\left(x^{2}+2 x z+12 x z^{2}\right)\right|_{0} ^{1} d y d z=\int_{0}^{3} \int_{0}^{2}\left(1+2 z+12 z^{2}\right) d y d z \\\&=\int_{0}^{3} 2\left(1+2 z+12 z^{2}\right) d z=\left.\left(2 z+2 z^{2}+8 z^{3}\right)\right|_{0} ^{3}=240\end{aligned}$$

\(\operatorname{div} \mathbf{F}=\frac{1}{x^{2}+y^{2}+z^{2}} .\) Using spherical coordinates $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D} \operatorname{div} \mathbf{F} d V=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{a}^{b} \frac{1}{\rho^{2}} \rho^{2} \sin \phi d \rho d \phi d \theta \\ &=\int_{0}^{2 \pi} \int_{0}^{\pi}(b-a) \sin \phi d \phi d \theta=(b-a) \int_{0}^{2 \pi}-\left.\cos \phi\right|_{0} ^{\pi} d \theta \\ &=(b-a) \int_{0}^{2 \pi} 2 d \theta=4 \pi(b-a)\end{aligned}$$

In the first octant, \(z=x+y\) is nonnegative. Then \(V=\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}}(x+y) d y d x=\left.\int_{0}^{3}\left(x y+\frac{1}{2} y^{2}\right)\right|_{0} ^{\sqrt{9-x^{2}}} d x\) \(=\int_{0}^{3}\left(x \sqrt{9-x^{2}}+\frac{9}{2}-\frac{1}{2} x^{2}\right) d x=\left.\left[-\frac{1}{3}\left(9-x^{2}\right)^{3 / 2}+\frac{9}{2} x-\frac{1}{6} x^{3}\right]\right|_{0} ^{3}=\left(\frac{27}{2}-\frac{9}{2}\right)-(-9)=18\).

$$\tan \theta=1 / \sqrt{3}, y / x=1 / \sqrt{3}, x=\sqrt{3} y, x > 0$$

\(\int_{0}^{2} \int_{y^{2}}^{4} \cos x^{3 / 2} d x d y=\int_{0}^{4} \int_{0}^{\sqrt{x}} \cos x^{3 / 2} d y d x=\left.\int_{0}^{4} y \cos x^{3 / 2}\right|_{0} ^{\sqrt{x}} d x\) \(=\int_{0}^{4} \sqrt{x} \cos x^{3 / 2} d x=\left.\frac{2}{3} \sin x^{3 / 2}\right|_{0} ^{4}=\frac{2}{3} \sin 8\)
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