$\text { Find } M _ { y } \text { for the lamina of uniform density } \rho \text { bounded by the graphs of the }$ equations $x = 4 y + 12$ and $x = y ^ { 2 }$.
A) $M _ { y } = \frac { 17,408 \rho } { 15 }$
B) $M _ { y } = \frac { 5,806 \rho } { 5 }$
C) $M _ { y } = \frac { 34,816 \rho } { 15 }$
D) $M _ { y } = \frac { 8,704 \rho } { 15 }$
E) $M _ { y } = \frac { 17,413 \rho } { 30 }$

Question

Quizplus · Accepted Answer

To find $M_y$, the moment about the y-axis, for the lamina, we integrate the product of the density, the distance from the y-axis (x in this case), and the differential area element. The area is bounded by $x = 4y + 12$ and $x = y^2$, so we first find the intersection points by setting $4y + 12 = y^2$, which simplifies to $y^2 - 4y - 12 = 0$. Solving this quadratic equation gives the y-values of the intersection points. Then, we set up the integral of $x \rho \, dy$ across the y-boundaries, with $x$ expressed as the difference between the two bounding functions. The correct calculation leads to $M_y = \frac{17,408 \rho}{15}$, indicating that choice A is correct.

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