Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 13: Problem 2
Evaluate the iterated integrals. $$\int_{0}^{\pi / 2} \int_{0}^{\sin \theta} r d r d \theta$$
Short Answer
Expert verified
The value of the iterated integral is \( \frac{\pi}{8} \).
Step by step solution
01
Evaluate the inner integral
Evaluate the inner integral \( \int_{0}^{\sin \theta} r \, d r \). This is an integration of \( r \) with respect to \( r \), so apply the power rule for integration: \[ \int r \, d r = \frac{r^2}{2} + C \].So, we obtain: \[ \int_{0}^{\sin \theta} r \, d r = \left[ \frac{r^2}{2} \right]_0^{\sin \theta} = \frac{(\sin \theta)^2}{2} - \frac{0^2}{2} = \frac{\sin^2 \theta}{2} \]
02
Set up the outer integral
Substitute the result from the inner integral into the outer integral. The expression now is: \[ \int_{0}^{\pi / 2} \frac{\sin^2 \theta}{2} \, d \theta \].
03
Simplify the outer integral
Factor out the constant from the integral: \[ \frac{1}{2} \int_{0}^{\pi / 2} \sin^2 \theta \, d \theta \]. Now we must integrate \( \sin^2 \theta \).
04
Use the identity to integrate
Use the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) to simplify the integral: \[ \frac{1}{2} \int_{0}^{\pi / 2} \frac{1 - \cos(2\theta)}{2} \, d\theta \]. This can be rewritten as: \[ \frac{1}{4} \int_{0}^{\pi / 2} 1 \, d\theta - \frac{1}{4} \int_{0}^{\pi / 2} \cos(2\theta) \, d\theta \].
05
Evaluate the integrals separately
Evaluate \( \frac{1}{4} \int_{0}^{\pi / 2} 1 \, d\theta \) and \( \frac{1}{4} \int_{0}^{\pi / 2} \cos(2\theta) \, d\theta \):For the first integral: \[ \frac{1}{4} \cdot \left[ \theta \right]_{0}^{\pi / 2} = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8} \].For the second integral, using substitution \( u = 2\theta \), \( du = 2 \, d\theta \) or \( d\theta = \frac{1}{2} du \):Convert the limits: when \( \theta = 0 \), \( u = 0 \); when \( \theta = \pi/2 \), \( u = \pi \).\[ \frac{1}{4} \cdot \frac{1}{2} \int_{0}^{\pi} \cos(u) \, du = \frac{1}{8} \left[ \sin(u) \right]_{0}^{\pi} = \frac{1}{8}(\sin(\pi) - \sin(0)) = \frac{1}{8}(0) = 0 \].
06
Combine the evaluated integrals
Add the evaluated results from the previous step to obtain the final answer: \[ \frac{\pi}{8} + 0 = \frac{\pi}{8} \].
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integration
Double integration is a powerful tool used in calculus to calculate the volume under a surface in a plane. In essence, it involves two successive integrations, also known as iterated integrals.
The idea is to first integrate with respect to one variable while treating the other as a constant, and then integrate the result with respect to the second variable.
For example, consider the double integral:
It also comes handy in statistics when finding probabilities over a certain region.
The idea is to first integrate with respect to one variable while treating the other as a constant, and then integrate the result with respect to the second variable.
For example, consider the double integral:
- First, evaluate the inner integral \( \int_{a}^{b} f(x, y) \, dx \), where \( x \) is the variable of integration.
- After finding this result, compute the outer integral \( \int_{c}^{d} \text{(result from inner integral)} \, dy \).
It also comes handy in statistics when finding probabilities over a certain region.
Polar Coordinates Integration
Polar coordinates offer a convenient way to deal with integration problems involving circular or radial symmetry, which are complex in Cartesian coordinates.
Instead of using \( x \) and \( y \), polar coordinates rely on \( r \), the radius, and \( \theta \), the angle.
To convert a function from Cartesian to polar, you use the transformations \( x = r \cos \theta \) and \( y = r \sin \theta \). The element of area \( dx \, dy \) becomes \( r \, dr \, d\theta \) in polar coordinates. This helps simplify many integrals, especially those involving circular regions.
When integrating iterated integrals in polar coordinates, as with the example:
Instead of using \( x \) and \( y \), polar coordinates rely on \( r \), the radius, and \( \theta \), the angle.
To convert a function from Cartesian to polar, you use the transformations \( x = r \cos \theta \) and \( y = r \sin \theta \). The element of area \( dx \, dy \) becomes \( r \, dr \, d\theta \) in polar coordinates. This helps simplify many integrals, especially those involving circular regions.
When integrating iterated integrals in polar coordinates, as with the example:
- First, integrate with respect to \( r \), keeping \( \theta \) constant. This usually addresses the radial aspect.
- Then compute the integral with respect to \( \theta \), covering the angular portion.
Trigonometric Substitution
Trigonometric substitution is a technique used in calculus to simplify integrals involving square roots and trigonometric expressions.
This method involves substituting a trigonometric function for a variable to transform a difficult integral into a more straightforward form.
For instance, if the integral contains something like \( \sqrt{a^2 - x^2} \), you can use the substitution \( x = a \sin \theta \), which simplifies the expression under the square root using a trigonometric identity.
Similarly, for expressions like \( \sqrt{x^2 + a^2} \) or \( \sqrt{x^2 - a^2} \), suitable substitutions can be \( x = a \tan \theta \) or \( x = a \sec \theta \) respectively.
Through trigonometric substitution:
This method involves substituting a trigonometric function for a variable to transform a difficult integral into a more straightforward form.
For instance, if the integral contains something like \( \sqrt{a^2 - x^2} \), you can use the substitution \( x = a \sin \theta \), which simplifies the expression under the square root using a trigonometric identity.
Similarly, for expressions like \( \sqrt{x^2 + a^2} \) or \( \sqrt{x^2 - a^2} \), suitable substitutions can be \( x = a \tan \theta \) or \( x = a \sec \theta \) respectively.
Through trigonometric substitution:
- The integral transforms into a form that involves simple trigonometric functions, making it easier to integrate.
- Once the integral is evaluated, you convert back to the original variable using inverse trigonometric functions.
