Consider the following. A = Convert the iterated integral to polar coordinates. A B S^L. ( Jo Jo B = √2x - x 1² √2²-²4. Jo || 472 4r 4√x² + y² dy dx X Evaluate the iterated integral. dr de 4
Consider the following. A = Convert the iterated integral to polar coordinates. A B S^L. ( Jo Jo B = √2x - x 1² √2²-²4. Jo || 472 4r 4√x² + y² dy dx X Evaluate the iterated integral. dr de 4
Calculus: Early Transcendentals
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Author:James Stewart
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![**Consider the following:**
\[ \int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} 4 \sqrt{x^2 + y^2} \, dy \, dx \]
**Convert the iterated integral to polar coordinates:**
\[ \int_{0}^{A} \int_{0}^{B} \left(4r^2\right) \, dr \, d\theta \]
- **A =** [Input field] (Incorrect, indicated with a red "X")
- **B =** [Input field] (Incorrect, indicated with a red "X")
**Evaluate the iterated integral.**
- [Input field]
**Need Help?** [Button: "Read It"]
[Button: "Submit Answer"]
Note: The incorrect inputs for \(A\) and \(B\) are marked with a red "X."
### Explanation:
To convert the given integral to polar coordinates, we need to express the Cartesian coordinates \( (x, y) \) in terms of polar coordinates \( (r, \theta) \). The transformation involves:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- The Jacobian of transformation \( \frac{\partial(x,y)}{\partial(r,\theta)} = r \).
This results in the double integral over polar coordinates for \( r \) and \( \theta \). The function \( \sqrt{x^2 + y^2} \) becomes \( r \) in polar coordinates. Thus, we integrate from \( r = 0 \) to the boundary in terms of \( \theta \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04750898-40b6-4507-9bb8-e18bccce9e61%2F95787728-4bdd-4ef9-9e16-91377b5802d4%2F6wpyw2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Consider the following:**
\[ \int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} 4 \sqrt{x^2 + y^2} \, dy \, dx \]
**Convert the iterated integral to polar coordinates:**
\[ \int_{0}^{A} \int_{0}^{B} \left(4r^2\right) \, dr \, d\theta \]
- **A =** [Input field] (Incorrect, indicated with a red "X")
- **B =** [Input field] (Incorrect, indicated with a red "X")
**Evaluate the iterated integral.**
- [Input field]
**Need Help?** [Button: "Read It"]
[Button: "Submit Answer"]
Note: The incorrect inputs for \(A\) and \(B\) are marked with a red "X."
### Explanation:
To convert the given integral to polar coordinates, we need to express the Cartesian coordinates \( (x, y) \) in terms of polar coordinates \( (r, \theta) \). The transformation involves:
- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- The Jacobian of transformation \( \frac{\partial(x,y)}{\partial(r,\theta)} = r \).
This results in the double integral over polar coordinates for \( r \) and \( \theta \). The function \( \sqrt{x^2 + y^2} \) becomes \( r \) in polar coordinates. Thus, we integrate from \( r = 0 \) to the boundary in terms of \( \theta \).
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