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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.6.60
Chapter 8, Problem 8.6.60

Moment about y-axis:
A thin plate of constant density δ = 1 occupies the region enclosed by the curve
y = 36/(2x + 3) and the line x = 3 in the first quadrant. Find the moment of the plate about the y-axis.

Verified step by step guidance
1
Identify the region and the density: The plate has constant density \( \delta = 1 \), and it occupies the region bounded by the curve \( y = \frac{36}{2x + 3} \), the vertical line \( x = 3 \), and the coordinate axes in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)).
Set up the expression for the moment about the y-axis: The moment about the y-axis, \( M_y \), for a lamina with density \( \delta \) is given by the integral \( M_y = \int x \cdot \delta \, dA \), where \( dA \) is the differential area element.
Express the differential area element \( dA \) in terms of \( dx \) and \( dy \): Since the region is bounded by \( y = \frac{36}{2x + 3} \), for each \( x \) between 0 and 3, the vertical slice has height \( y \). So, \( dA = y \, dx = \frac{36}{2x + 3} \, dx \).
Write the integral for the moment about the y-axis: Substitute \( dA \) and \( \delta = 1 \) into the moment formula to get \( M_y = \int_0^3 x \cdot 1 \cdot \frac{36}{2x + 3} \, dx = \int_0^3 \frac{36x}{2x + 3} \, dx \).
Evaluate the integral: To solve \( \int_0^3 \frac{36x}{2x + 3} \, dx \), consider using substitution or algebraic manipulation (such as dividing numerator and denominator or rewriting the integrand) to simplify the integral before integrating.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment about the y-axis

The moment about the y-axis measures how the mass of a plate is distributed relative to the y-axis. It is calculated by integrating the product of the x-coordinate and the mass element over the region. For a plate with density δ, the moment is M_y = ∫∫ x δ dA, where dA is the differential area element.
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Setting up the region of integration

To find the moment, the region bounded by the curve y = 36/(2x + 3), the line x = 3, and the coordinate axes must be clearly defined. This involves determining the limits for x and y, ensuring the integration covers the entire plate in the first quadrant. Proper limits are essential for accurate calculation.
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Double integration for area and moments

Double integrals allow calculation of quantities over two-dimensional regions. Here, integrating δ x dA over the plate's area sums the contributions of all small elements. The integral is often set up as ∫ from x=0 to 3 ∫ from y=0 to y=36/(2x+3) of x δ dy dx, simplifying the evaluation of the moment.
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