Find the mass of the lamina that occupies the region $D = \left\{ ( x , y ) \mid 0 \leq x \leq 1 , x ^ { 2 } \leq y \leq 1 \right\}$ and has density function p(x, y) = x + y.
A)0.35
B)0.40
C)0.45
D)0.50
E)0.55
F)0.60
G)0.65
H)0.70

Question

Quizplus · Accepted Answer

The mass of the lamina can be found by integrating the density function over the given region $D$. The integral for the mass $m$ is given by:$$m = \int \int_D (x + y) \, dy \, dx$$Given the bounds for $D$, we integrate $y$ from $x^2$ to $1$ and $x$ from $0$ to $1$:$$m = \int_0^1 \int_{x^2}^1 (x + y) \, dy \, dx$$First, integrate with respect to $y$:$$= \int_0^1 \left[ xy + \frac{y^2}{2} \right]_{x^2}^1 \, dx$$$$= \int_0^1 \left( x + \frac{1}{2} - x^3 - \frac{x^4}{2} \right) \, dx$$$$= \int_0^1 \left( x + \frac{1}{2} - x^3 - \frac{x^4}{2} \right) \, dx$$$$= \left[ \frac{x^2}{2} + \frac{x}{2} - \frac{x^4}{4} - \frac{x^5}{10} \right]_0^1$$$$= \frac{1}{2} + \frac{1}{2} - \frac{1}{4} - \frac{1}{10}$$$$= 1 - \frac{1}{4} - \frac{1}{10}$$$$= \frac{10}{10} - \frac{2.5}{10} - \frac{1}{10}$$$$= \frac{10 - 2.5 - 1}{10}$$$$= \frac{6.5}{10}$$$$= 0.65$$Therefore, the mass of the lamina is 0.65.

Find the Mass of the Lamina That Occupies the Region D={(x,y)∣0≤x≤1,x2≤y≤1}D = \left\{ ( x , y ) \mid 0 \leq x \leq 1 , x ^ { 2 } \leq y \leq 1 \right\}D={(x,y)∣0≤x≤1,x2≤y≤1}

Find the Mass of the Lamina That Occupies the Region $D = \left\{ ( x , y ) \mid 0 \leq x \leq 1 , x ^ { 2 } \leq y \leq 1 \right\}$