Find the mass of the lamina that occupies the region $D$ and has the given density function, if $D$ is bounded by the parabola $x = y ^ { 2 }$ and the line $y = x - 2$.
$$\rho ( x , y ) = 3$$
Select the correct answer.
A) $\frac { 3 } { 2 }$
B) 2
C) 27
D) $\frac { 27 } { 2 }$
E) None of these

Question

Quizplus · Accepted Answer

To find the mass of the lamina, we use the formula:

$$M = \iint\limits_D ho(x,y)\ dA$$

where $D$ is the region of integration and $ho(x,y)$ is the density.

The region $D$ is bounded by the parabola $x = y^2$ and the line $y=x-2$. We can rewrite the line as $y-x=-2$, and solve the system of equations to find the points of intersection:

\begin{align*} x &= y^2 \ y-x &= -2 \end{align*}

Adding the equations gives $y^2-y-2=0$, which factors as $(y-2)(y+1)=0$. Thus, $y=-1$ or $y=2$. Substituting into $x=y^2$, we find $x=1$ or $x=4$, respectively. Therefore, $D$ is the region bounded by the curves $y=x-2$, $y=-1$, and $x=y^2$, for $1\leq x\leq 4$.

Now we can set up the integral:

$$ M = \iint\limits_D ho(x,y)\ dA = \iint\limits_D 3\ dA $$

We can either integrate with respect to $x$ first or with respect to $y$ first. Since the limits of integration are easier to write with respect to $y$, we integrate with respect to $y$ first:

\begin{align*} M &= \int\limits_{-1}^2 \int\limits_{y^2}^{y+2} 3\ dx\ dy \ &= \int\limits_{-1}^2 3(y+2-y^2)\ dy \ &= \left[3\left(\frac{y^2}{2}+2y-\frac{y^3}{3}ight)ight]_{-1}^2 \ &= \frac{27}{2} \end{align*}

Therefore, the mass of the lamina that occupies $D$ with density function $ho(x,y)=3$ is $\frac{27}{2}$. The answer is (D).

Find the Mass of the Lamina That Occupies the Region DDD

Find the Mass of the Lamina That Occupies the Region $D$