Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.CR: Chapter 8 Review
Problem 11CR
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Question
![To evaluate the given triple integral:
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \int_{0}^{\frac{2}{y}} 8 \sin y \, dz \, dx \, dy. \]
We need to carefully consider the order of integration and integrate step by step:
1. Integrate with respect to \( z \):
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \left[ \int_{0}^{\frac{2}{y}} 8 \sin y \, dz \right] dx \, dy. \]
2. The integral inside can be evaluated as:
\[ \int_{0}^{\frac{2}{y}} 8 \sin y \, dz = 8 \sin y \left. z \right|_{0}^{\frac{2}{y}} = 8 \sin y \left( \frac{2}{y} - 0 \right) = \frac{16 \sin y}{y}. \]
3. Now, the triple integral reduces to:
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \frac{16 \sin y}{y} \, dx \, dy. \]
4. Integrate with respect to \( x \):
\[ \int_{0}^{\frac{\pi}{2}} \left[ \int_{0}^{2y} \frac{16 \sin y}{y} \, dx \right] dy = \int_{0}^{\frac{\pi}{2}} \frac{16 \sin y}{y} \left. x \right|_{0}^{2y} \, dy = \int_{0}^{\frac{\pi}{2}} \frac{16 \sin y}{y} \cdot 2y \, dy = \int_{0}^{\frac{\pi}{2}} 32 \sin y \, dy. \]
5. Finally, integrate with respect to \( y \):
\[ 32 \int_{0}^{\frac{\pi}{2}} \sin y \, dy. \]
We know that:
\[ \int \sin y \, dy = -\cos y. \]
So,
\[ 32 \left. -\cos y \right|_{0}^{\frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1390cd79-ed02-4e3f-b0a6-b6020cd524ef%2F7e598e05-4523-4667-ab03-aa0936829c11%2F0bnttt5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To evaluate the given triple integral:
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \int_{0}^{\frac{2}{y}} 8 \sin y \, dz \, dx \, dy. \]
We need to carefully consider the order of integration and integrate step by step:
1. Integrate with respect to \( z \):
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \left[ \int_{0}^{\frac{2}{y}} 8 \sin y \, dz \right] dx \, dy. \]
2. The integral inside can be evaluated as:
\[ \int_{0}^{\frac{2}{y}} 8 \sin y \, dz = 8 \sin y \left. z \right|_{0}^{\frac{2}{y}} = 8 \sin y \left( \frac{2}{y} - 0 \right) = \frac{16 \sin y}{y}. \]
3. Now, the triple integral reduces to:
\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{2y} \frac{16 \sin y}{y} \, dx \, dy. \]
4. Integrate with respect to \( x \):
\[ \int_{0}^{\frac{\pi}{2}} \left[ \int_{0}^{2y} \frac{16 \sin y}{y} \, dx \right] dy = \int_{0}^{\frac{\pi}{2}} \frac{16 \sin y}{y} \left. x \right|_{0}^{2y} \, dy = \int_{0}^{\frac{\pi}{2}} \frac{16 \sin y}{y} \cdot 2y \, dy = \int_{0}^{\frac{\pi}{2}} 32 \sin y \, dy. \]
5. Finally, integrate with respect to \( y \):
\[ 32 \int_{0}^{\frac{\pi}{2}} \sin y \, dy. \]
We know that:
\[ \int \sin y \, dy = -\cos y. \]
So,
\[ 32 \left. -\cos y \right|_{0}^{\frac
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