/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 \(I_{y}=\int_{0}^{4} \int_{0}^{\... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 55

\(I_{y}=\int_{0}^{4} \int_{0}^{\sqrt{y}} x^{2} y d x d y=\left.\int_{0}^{4} \frac{1}{3} x^{3} y\right|_{0} ^{\sqrt{y}} d y=\frac{1}{3} \int_{0}^{4} y^{3 / 2} y d y=\frac{1}{3} \int_{0}^{4} y^{5 / 2} d y\) \(=\left.\frac{1}{3}\left(\frac{2}{7} y^{7 / 2}\right)\right|_{0} ^{4}=\frac{2}{21}\left(4^{7 / 2}\right)=\frac{256}{21}\)

Short Answer

Expert verified
The value of the integral \( I_y \) is \( \frac{256}{21} \).

Step by step solution

01

Set up the double integral

The goal is to solve the double integral \( I_y = \int_{0}^{4} \int_{0}^{\sqrt{y}} x^{2} y \, dx \, dy \). The limits of integration indicate that we first integrate with respect to \( x \) from 0 to \( \sqrt{y} \), and then with respect to \( y \) from 0 to 4.
02

Integrate with respect to x

Begin by integrating the function \( x^2 y \) with respect to \( x \), keeping \( y \) as a constant. The antiderivative of \( x^2 \) is \( \frac{1}{3} x^3 \). So, we get:\[ \int x^2 y \, dx = y \cdot \frac{1}{3} x^3 = \frac{1}{3} y x^3 \]We need to evaluate this from \( x=0 \) to \( x=\sqrt{y} \).
03

Evaluate the inner integral

Substitute the limits \( x = 0 \) and \( x = \sqrt{y} \) into the antiderivative:\[ \left. \frac{1}{3} y x^3 \right|_{0}^{\sqrt{y}} = \frac{1}{3} y (\sqrt{y})^3 - \frac{1}{3} y (0)^3 = \frac{1}{3} y \cdot y^{3/2} = \frac{1}{3} y^{5/2} \]This simplifies the integral to \( \frac{1}{3} y^{5/2} \).
04

Integrate with respect to y

Now integrate \( \int_{0}^{4} \frac{1}{3} y^{5/2} \, dy \). The antiderivative of \( y^{5/2} \) is \( \frac{2}{7} y^{7/2} \). Therefore, we have:\[ \frac{1}{3} \int y^{5/2} \, dy = \frac{1}{3} \left( \frac{2}{7} y^{7/2} \right) = \frac{2}{21} y^{7/2} \]Evaluate from \( y = 0 \) to \( y = 4 \).
05

Evaluate the outer integral

Substitute the limits \( y = 0 \) and \( y = 4 \) into the antiderivative:\[ \frac{2}{21} \left. y^{7/2} \right|_{0}^{4} = \frac{2}{21} \left( 4^{7/2} - 0^{7/2} \right) \]Where \( 4^{7/2} = (2^2)^{7/2} = 2^7 = 128 \). Hence,\[ \frac{2}{21} \times 128 = \frac{256}{21} \]
06

Conclusion

The value of the double integral \( I_y \) is \( \frac{256}{21} \). This corresponds to the area under the defined function across the specified region.

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Key Concepts

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

Integration Techniques
When working with double integrals, understanding how to apply integration techniques can significantly simplify the process. Double integrals involve performing two integrations: one for each variable. The integration is done over a two-dimensional region. In this particular problem, the integration is over variables \( x \) and \( y \), with specific limits.
  • First, identify which variable to integrate first. Often, this is determined by the order given in the integral. Here, \( x \) is integrated first from 0 to \( \sqrt{y} \).
  • Second, perform the integration with respect to \( x \), treating \( y \) as a constant. This technique is similar in concept to single variable calculus but in a nested manner.
  • Lastly, substitute the bounds for the first variable and integrate with respect to the remaining variable. This two-step process is essential for tackling double integrals.
Effective use of integration techniques enables you to break down complex integral functions into manageable steps, making the process less daunting.
Definite Integrals
Definite integrals are a fundamental concept when working with continuous functions to find the total quantity of interest, like area, in a specified interval. For double integrals, their primary purpose is to compute the volume under a surface over a specified two-dimensional area.
  • The specific limits for each integral define the region over which the function is integrated. Here, we have \( x \) ranging from 0 to \( \sqrt{y} \) and \( y \) ranging from 0 to 4.
  • The process involves evaluating the antiderivative at the upper limit and subtracting the antiderivative evaluated at the lower limit.
  • In our exercise, definite integration ensures that we are accurately measuring the quantity represented by the integral, concluding with a specific numerical result, \( \frac{256}{21} \) in this example.
Through definite integrals, we can derive exact values representing physical quantities, effectively linking abstract calculus concepts to real-world applications.
Calculus Applications
Calculus, particularly through the use of double integrals, finds extensive applications across various fields such as physics, engineering, and economics. It plays a crucial role in visualizing and calculating multi-dimensional data.
  • Double integrals can be applied to compute areas, volumes, centroids, and other significant physical quantities over surfaces and regions.
  • In this exercise, by applying double integrals, we calculated the value \( \frac{256}{21} \), representing the volume under the defined surface across the given limits.
  • Comprehension of double integrals in calculus equips you with tools to solve practical problems involving complex geometric shapes and distributions.
By mastering calculus applications, you not only solve textbook problems but also develop analytical skills applicable in real-world scenarios.

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Most popular questions from this chapter

We are given \(\rho(x, y, z)=k z\). $$\begin{aligned} m &=\int_{0}^{8} \int_{0}^{4} \int_{0}^{y^{1 / 3}} k z d x d z d y=\left.k \int_{0}^{8} \int_{0}^{4} x z\right|_{0} ^{y^{1 / 3}} d z d y=k \int_{0}^{8} \int_{0}^{4} y^{1 / 3} z d z d y \\ &=\left.k \int_{0}^{8} \frac{1}{2} y^{1 / 3} z^{2}\right|_{0} ^{1} d y=8 k \int_{0}^{8} y^{1 / 3} d y=\left.8 k\left(\frac{3}{4} y^{4 / 3}\right)\right|_{0} ^{8}=96 k \end{aligned}$$ $$\begin{aligned} M_{x z} &=\int_{0}^{8} \int_{0}^{4} \int_{0}^{y^{1 / 3}} k y z d x d z d y=\left.k \int_{0}^{8} \int_{0}^{4} x y z\right|_{0} ^{y^{1 / 3}} d z d y=k \int_{0}^{8} \int_{0}^{4} y^{4 / 3} z d z d y \\ &=\left.k \int_{0}^{8} \frac{1}{2} y^{4 / 3} z^{2}\right|_{0} ^{4} d y=8 k \int_{0}^{8} y^{4 / 3} d y=\left.8 k\left(\frac{3}{7} y^{7 / 3}\right)\right|_{0} ^{8}=\frac{3072}{7} k \\ M_{y z} &=\int_{0}^{8} \int_{0}^{4} \int_{0}^{y^{1 / 3}} k x z d x d z d y=\left.k \int_{0}^{8} \int_{0}^{4} \frac{1}{2} x^{2} z\right|_{0} ^{y^{1 / 3}} d z d y=\frac{1}{2} k \int_{0}^{8} \int_{0}^{4} y^{2 / 3} z d z d y \\ &=\left.\frac{1}{2} k \int_{0}^{8} \frac{1}{2} y^{2 / 3} z^{2}\right|_{0} ^{4} d y=4 k \int_{0}^{8} y^{2 / 3} d y=\left.4 k\left(\frac{3}{5} y^{5 / 3}\right)\right|_{0} ^{8}=\frac{384}{5} k \end{aligned}$$ $$\bar{x}=M_{y z} / m=\frac{384 k / 5}{96 k}=4 / 5 ; \bar{y}=M_{x z} / m=\frac{3072 k / 7}{96 k}=32 / 7 ; \bar{z}=M_{x y} / m=\frac{256 k}{96 k}=8 / 3$$ The center of mass is \((4 / 5,32 / 7,8 / 3)\).

$$\mathbf{F}=t^{3} \mathbf{i}+t^{4} \mathbf{j}+t^{5} \mathbf{k} ; \quad d \mathbf{r}=3 t^{2} \mathbf{i}+2 t \mathbf{j}+\mathbf{k} ; \quad W=\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{1}^{3}\left(3 t^{5}+2 t^{5}+t^{5}\right) d t=\int_{1}^{3} 6 t^{5} d t=\left.t^{6}\right|_{1} ^{3}=728$$

since the height of the triangle is \(x \sin \theta,\) the area is given by \(A=\frac{1}{2} x y \sin \theta .\) Then $$\frac{d A}{d t} \frac{\partial A}{\partial x} \frac{d x}{d t}+\frac{\partial A}{\partial y} \frac{d y}{d t}+\frac{\partial A}{\partial \theta} \frac{d \theta}{d t}=\frac{1}{2} y \sin \theta \frac{d x}{d t}+\frac{1}{2} x \sin \theta \frac{d y}{d t}+\frac{1}{2} x y \cos \theta \frac{d \theta}{d t}.$$ When \(x=10, y=8, \theta=\pi / 6, d x / d t=0.3, d y / d t=0.5,\) and \(d \theta / d t=0.1,\) $$\begin{aligned}\frac{d A}{d t} &=\frac{1}{2}(8)\left(\frac{1}{2}\right)(0.3)+\frac{1}{2}(10)\left(\frac{1}{2}\right)(0.5)+\frac{1}{2}(10)(8)\left(\frac{\sqrt{3}}{2}\right)(0.1) \\\&=0.6+1.25+2 \sqrt{3}=1.85+2 \sqrt{3} \approx 5.31 \mathrm{cm}^{2} / \mathrm{s}.\end{aligned}$$

Using symmetry, \(I_{x}=2 \int_{0}^{\pi / 2} \int_{0}^{\cos x} k y^{2} d y d x=\left.2 k \int_{0}^{\pi / 2} \frac{1}{3} y^{3}\right|_{0} ^{\cos x} d x=\frac{2}{3} k \int_{0}^{\pi / 2} \cos ^{3} x d x\) \(=\frac{2}{3} k \int_{0}^{\pi / 2} \cos x\left(1-\sin ^{2} x\right) d x=\left.\frac{2}{3} k\left(\sin x-\frac{1}{3} \sin ^{3} x\right)\right|_{0} ^{\pi / 2}=\frac{4}{9} k\).

The surface is \(g(x, y, z)=x^{2}+y^{2}+z-5=0 . \nabla g=2 x \mathbf{i}+2 y \mathbf{j}+\mathbf{k}\) \(|\nabla g|=\sqrt{1+4 x^{2}+4 y^{2}} ; \quad \mathbf{n}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+\mathbf{k}}{\sqrt{1+4 x^{2}+4 y^{2}}} ; \mathbf{F} \cdot \mathbf{n}=\frac{z}{\sqrt{1+4 x^{2}+4 y^{2}}}\) \(z_{x}=-2 x, z_{y}=-2 y, d S=\sqrt{1+4 x^{2}+4 y^{2}} d A .\) Using polar coordinates $$\begin{aligned} \mathrm{Flux} &=\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=\iint_{R} \frac{z}{\sqrt{1+4 x^{2}+4 y^{2}}} \sqrt{1+4 x^{2}+4 y^{2}} d A=\iint_{R}\left(5-x^{2}-y^{2}\right) d A \\ &=\int_{0}^{2 \pi} \int_{0}^{2}\left(5-r^{2}\right) r d r d \theta=\left.\int_{0}^{2 \pi}\left(\frac{5}{2} r^{2}-\frac{1}{4} r^{4}\right)\right|_{0} ^{2} d \theta=\int_{0}^{2 \pi} 6 d \theta=12 \pi \end{aligned}$$
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