Question

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

Accepted Answer

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.

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.

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