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Chapter 5: Problem 12

Compute \(\iint_{D} f(x, y) d A,\) where \(f(x, y)=y^{2} \sqrt{x}\) and \(D\) is the set of \((x, y),\) where \(x>0, y>x^{2},\) and \(y<10-x^{2}\)

Short Answer

Expert verified
Calculate using iterated integration from \( x = 0 \) to \( \sqrt{5} \) and \( y = x^2 \) to \( y = 10-x^2 \).

Step by step solution

01

Examine the region D

The region \( D \) is defined by the inequalities: \( x > 0 \), \( y > x^2 \), and \( y < 10 - x^2 \). This describes a region bounded by the parabolas \( y = x^2 \) and \( y = 10 - x^2 \). The intersection points of these curves can be found by setting \( x^2 = 10 - x^2 \).
02

Find bounds for x

Solve the equation \( x^2 = 10 - x^2 \) to find the x-values where the parabolas intersect. This simplifies to \( 2x^2 = 10 \), giving \( x^2 = 5 \) and thus \( x = \sqrt{5} \). Since \( x > 0 \), our x bounds are from \( 0 \) to \( \sqrt{5} \).
03

Find bounds for y

For each fixed \( x \), the region \( D \) is between \( y = x^2 \) and \( y = 10 - x^2 \). Thus, for each \( x \) in the interval \( (0, \sqrt{5}) \), \( y \) will range from \( x^2 \) to \( 10 - x^2 \).
04

Set up the iterated integral

The integral \( \iint_D f(x,y) \, dA \) is set up as \( \int_0^{\sqrt{5}} \int_{x^2}^{10-x^2} y^2 \sqrt{x} \, dy \, dx \).
05

Integrate with respect to y

First, compute the inner integral \( \int_{x^2}^{10-x^2} y^2 \, dy \). The antiderivative is \( \frac{y^3}{3} \). Evaluate this from \( y = x^2 \) to \( y = 10 - x^2 \) to get: \[ \left( \frac{(10-x^2)^3}{3} \right) - \left( \frac{(x^2)^3}{3} \right). \]
06

Integrate with respect to x

Substitute the result of the inner integral into the outer integral: \( \int_0^{\sqrt{5}} \sqrt{x} \left( \frac{(10-x^2)^3}{3} - \frac{(x^2)^3}{3} \right) \, dx \). Simplify and compute this integral to find the total value.

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Key Concepts

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

Region of Integration
To solve a double integral, it鈥檚 crucial to understand the **region of integration**.The region, denoted as \(D\), is where we want to compute the integral.For this problem, the region \(D\) is defined by the conditions \(x > 0\), \(y > x^2\), and \(y < 10 - x^2\).This means we are working within boundaries created by two parabolas.
To visualize this, picture the Cartesian plane:
  • The parabola \(y = x^2\) opens upwards and starts at the origin.
  • The parabola \(y = 10 - x^2\) is a downward-facing parabola, looping down from the y-axis before crossing back.
The region \(D\) lies between these two parabolic curves, with \(x\) running from **0** on the left, and constrained by the intersection points of these curves on the right.
Iterated Integral
An **iterated integral** allows us to evaluate a double integral using a specific order of integration.It simplifies the process by breaking it down into two single integrals.In this exercise, we set up the iterated integral \[\int_0^{\sqrt{5}} \int_{x^2}^{10-x^2} y^2 \sqrt{x} \, dy \, dx \]This shows we first integrate with respect to \(y\), and then \(x\).
When dealing with iterated integrals:
  • Start with the innermost integral (here, integrate over \(y\)) to collapse the area into a single line along the other variable (\(x\)).
  • The next step considers the effect along the second variable by integrating the result.
Specify the bounds based on the region of integration鈥攈ere, found from parabolas' equations鈥攚hich are \(y = x^2\) to \(y = 10-x^2\), and those of \(x\) from \(0\) to \(\sqrt{5}\).
Parabola Intersection
**Parabola intersection** is essential to understand when defining the limits of integration for some regions.To find where two parabolas intersect, solve the equation from setting their expressions equal: \(x^2 = 10 - x^2\).
This reveals the intersection points:
  • Rearrange the equation to \(2x^2 = 10\), leading to \(x^2 = 5\).
  • Thus, \(x = \sqrt{5}\) (only \(x > 0\) applies due to the region condition).
The intersection points are where the region of integration ends along the \(x\)-axis.This information helps in setting the limits for our outer integral when evaluating the double integral.The curves split the plane into sections, and we're focused on the region between them.

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Most popular questions from this chapter

Find the volume of the region in Exercises 11 to 14. The region common to the intersecting cylinders \(x^{2}+y^{2} \leq a^{2}\) and \(x^{2}+z^{2} \leq a^{2}.\)

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Evaluate over the region \(R\): $$\iint_{R} \frac{y x^{3}}{y^{2}+2} d y d x, \quad R:[0,2] \times[-1,1]$$
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