Find the mass of the region bounded by y = x2, the y axis and the horizontal line y = 4 for x > 0 if the density function p(x,y) = x° y. Round your answer to four decimal places. %3D Mass 17
Find the mass of the region bounded by y = x2, the y axis and the horizontal line y = 4 for x > 0 if the density function p(x,y) = x° y. Round your answer to four decimal places. %3D Mass 17
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Problem Statement:**
Find the mass of the region bounded by \( y = x^2 \), the y-axis, and the horizontal line \( y = 4 \) for \( x \geq 0 \) if the density function \( \rho(x,y) = x^6 y \). Round your answer to four decimal places.
**Solution:**
To find the mass \( M \) of the region, use the double integral:
\[
M = \iint_R \rho(x,y) \, dA
\]
where \( R \) is the region described.
**Bounds:**
1. **Horizontal Boundaries:**
- From the problem, we have the curve \( y = x^2 \) and the line \( y = 4 \).
- This implies that \( x \) ranges from 0 to 2 because \( y = x^2 = 4 \Rightarrow x = \sqrt{4} = 2 \).
2. **Vertical Boundaries:**
- For a given \( x \), \( y \) ranges from \( x^2 \) to 4.
**Integral Setup:**
\[
M = \int_{0}^{2} \int_{x^2}^{4} x^6 y \, dy \, dx
\]
**Calculate the Integral:**
1. Integrate with respect to \( y \):
\[
\int_{x^2}^{4} x^6 y \, dy = \left[ \frac{x^6 y^2}{2} \right]_{x^2}^{4}
\]
\[
= \frac{x^6 (4)^2}{2} - \frac{x^6 (x^2)^2}{2}
\]
\[
= \frac{16x^6}{2} - \frac{x^{10}}{2}
\]
\[
= 8x^6 - \frac{x^{10}}{2}
\]
2. Integrate with respect to \( x \):
\[
\int_{0}^{2} \left( 8x^6 - \frac{x^{10}}{2} \right) \, dx
\]
3. Evaluate the integral:
\[
= \int_{0}^{2} 8x^6 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a1263ae-327d-4428-a0aa-137e6771516c%2F2271404e-0cb7-4a4b-bd5e-6ebc73c227a5%2Fcb7lk3_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the mass of the region bounded by \( y = x^2 \), the y-axis, and the horizontal line \( y = 4 \) for \( x \geq 0 \) if the density function \( \rho(x,y) = x^6 y \). Round your answer to four decimal places.
**Solution:**
To find the mass \( M \) of the region, use the double integral:
\[
M = \iint_R \rho(x,y) \, dA
\]
where \( R \) is the region described.
**Bounds:**
1. **Horizontal Boundaries:**
- From the problem, we have the curve \( y = x^2 \) and the line \( y = 4 \).
- This implies that \( x \) ranges from 0 to 2 because \( y = x^2 = 4 \Rightarrow x = \sqrt{4} = 2 \).
2. **Vertical Boundaries:**
- For a given \( x \), \( y \) ranges from \( x^2 \) to 4.
**Integral Setup:**
\[
M = \int_{0}^{2} \int_{x^2}^{4} x^6 y \, dy \, dx
\]
**Calculate the Integral:**
1. Integrate with respect to \( y \):
\[
\int_{x^2}^{4} x^6 y \, dy = \left[ \frac{x^6 y^2}{2} \right]_{x^2}^{4}
\]
\[
= \frac{x^6 (4)^2}{2} - \frac{x^6 (x^2)^2}{2}
\]
\[
= \frac{16x^6}{2} - \frac{x^{10}}{2}
\]
\[
= 8x^6 - \frac{x^{10}}{2}
\]
2. Integrate with respect to \( x \):
\[
\int_{0}^{2} \left( 8x^6 - \frac{x^{10}}{2} \right) \, dx
\]
3. Evaluate the integral:
\[
= \int_{0}^{2} 8x^6 \
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