Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 13: Problem 2
Evaluate each of the iterated integrals. $$ \int_{-2}^{2} \int_{0}^{1}\left(9-x^{2}\right) d y d x $$
Short Answer
Expert verified
The evaluated integral is \( \frac{92}{3} \).
Step by step solution
01
Understand the Problem Statement
We're given a double integral, \( \int_{-2}^{2} \int_{0}^{1}(9-x^{2}) \, dy \, dx \), which involves integrating the function \((9-x^2)\) first with respect to \(y\) and then with respect to \(x\). We'll solve this iterated integral step by step.
02
Integrate with Respect to y
Start by integrating the inner integral \( \int_{0}^{1}(9-x^{2}) \, dy \). Since \(9 - x^2\) is independent of \(y\), the integral simplifies to:\[\int_{0}^{1} (9-x^2) \, dy = (9-x^2) \int_{0}^{1} \, dy = (9-x^2)[y]_{0}^{1}\]Evaluating from 0 to 1, we get:\[(9-x^2)(1 - 0) = 9-x^2\]
03
Integrate with Respect to x
Now, we integrate the result from the previous step with respect to \(x\):\[\int_{-2}^{2} (9-x^2) \, dx\]This breaks down as:\[\int_{-2}^{2} 9 \, dx - \int_{-2}^{2} x^2 \, dx\]Calculate each integral separately.
04
Calculate the Integral of Constant 9
For the integral \( \int_{-2}^{2} 9 \, dx \):\[9 \int_{-2}^{2} \, dx = 9[x]_{-2}^{2} = 9(2 - (-2)) = 9 \times 4 = 36\]
05
Calculate the Integral of \(x^2\)
For the integral \( \int_{-2}^{2} x^2 \, dx \):\[\int x^2 \, dx = \frac{x^3}{3} \]Evaluate from -2 to 2:\[\left[\frac{x^3}{3}\right]_{-2}^{2} = \left(\frac{2^3}{3}\right) - \left(\frac{(-2)^3}{3}\right) = \frac{8}{3} - \left(-\frac{8}{3}\right)\]\[\frac{8}{3} + \frac{8}{3} = \frac{16}{3}\]
06
Combine the Results
Subtract the results from Step 4 and Step 5:\[36 - \frac{16}{3}\]Convert 36 to a fraction to combine:\[\frac{108}{3} - \frac{16}{3} = \frac{92}{3}\]
07
Final Answer
The evaluated value of the iterated integral is \( \frac{92}{3} \).
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.
Iterated Integrals
Iterated integrals are a fundamental concept in calculus for calculating the area, volume, or other accumulations over a region by performing integration in stages. The idea is simple: we evaluate a function by first integrating with respect to one variable, and then we proceed to integrate the result with respect to another.
- The inner integral is evaluated first, holding the outer variable constant, which simplifies the initial expression.
- The result of this integration is then used as the function for the outer integral.
Integration with Respect to a Variable
Integration with respect to a variable means considering that variable alone as the one changing, while treating others like constants. This is key in handling multiple integrals, like the ones in the original exercise.
- For the inner integral \( \int_{0}^{1}(9-x^2) \, dy \), \( x \) is a constant, so the expression simplifies.
- This results in the straightforward integration of a constant, such as \( (9-x^2) \times [y]_0^1 \).
Combining Integral Results
Once both integrals are evaluated, it is time to combine the results. This involves arithmetic operations to bring together the answers from each step.
- The separate integration outcomes are often combined through addition or subtraction, as seen in the derived expressions like \( \int 9 \, dx \) and \( \int x^2 \, dx \).
- From the exercise, the combinations of \( 36 \) and \( \frac{16}{3} \) are achieved by finding a common denominator for fractions, resulting in \( \frac{92}{3} \).
