Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.

D is bounded by
y =sqrt(x), y = 0, and x = 1; ρ(x, y) = 27x

m=?
(x,y)=?

consider the lamina as a collection of thin vertical strips, of height y and width dx. Just add up all the masses.

Each strip's mass is its area times it density. If we integrate along x, the density is a constant for each strip.

m = ∫[0,1] ρ y dx
= ∫[0,1] 27x √x dx
= 54/5 x^(5/2) [0,1]
= 54/5

To find the mass and center of mass of the lamina, we will integrate the density function over the region D.

To find the mass (m), we integrate the density function ρ(x, y) over the region D:

m = ∬ρ(x, y) dA

Here, dA represents the differential area element.

D is defined by the following boundaries:
y = sqrt(x), y = 0, and x = 1

Thus, the integral for the mass (m) is given by:

m = ∫∫ρ(x, y) dA

To evaluate this integral, we need to set up the limits of integration. Since the region D is bounded by y = sqrt(x), y = 0, and x = 1, we can set up the integral as follows:

m = ∫[0, 1] ∫[0, sqrt(x)] 27x dy dx

Now, let's evaluate this integral:

m = ∫[0, 1] [27x * y] [0, sqrt(x)] dx
= ∫[0, 1] 27x * sqrt(x) dx

To find the center of mass (x, y), we need to calculate the moments with respect to y and x, respectively, and then divide them by the total mass.

The moment with respect to y is given by:

Mx = ∬y * ρ(x, y) dA

And the moment with respect to x is given by:

My = ∬x * ρ(x, y) dA

For the center of mass (x, y), we can use the following formulas:

x = Mx / m
y = My / m

Let's calculate the moments with respect to y (Mx) and x (My), and then we can find the center of mass (x, y).

To find the mass (m) and center of mass of a lamina with a given density function ρ, you'll need to perform a double integration over the region D.

First, let's find the limits of integration for x and y.

The region D is bounded by y = √(x), y = 0, and x = 1. Since y is bounded by √(x) and 0, we know that y ranges from 0 to √(x). And x ranges from 0 to 1.

Now, let's calculate the mass (m). The mass of the lamina can be found by integrating the density function ρ(x, y) over the region D. The density function given is ρ(x, y) = 27x.

Therefore, the mass is given by the double integral of 27x over the region D:

m = ∫∫D ρ(x, y) dA

m = ∫∫D 27x dA

where dA represents the infinitesimal area element.

We can rewrite the double integral in terms of x and y as follows:

m = ∫[0,1]∫[0,√(x)] 27x dy dx

Now, let's evaluate this double integral step by step.

First, we integrate with respect to y, treating x as a constant:

m = ∫[0,1] 27x [y]dy evaluated from y = 0 to y = √(x)

m = ∫[0,1] 27x (√(x) - 0) dx

m = ∫[0,1] 27x^(3/2) dx

Now, we integrate with respect to x:

m = [18x^(5/2)] evaluated from x = 0 to x = 1

m = 18(1)^(5/2) - 18(0)^(5/2)

m = 18(1) - 18(0)

m = 18

So, the mass of the lamina is 18.

Now, let's find the center of mass (x̄, ȳ) of the lamina. The center of mass is given by:

x̄ = (1/m) ∫∫D x ρ(x, y) dA
ȳ = (1/m) ∫∫D y ρ(x, y) dA

To find x̄, we integrate x times the density function ρ over the region D and divide by the mass m:

x̄ = (1/18) ∫∫D x(27x) dA

Similarly, to find ȳ, we integrate y times the density function ρ over the region D and divide by the mass m:

ȳ = (1/18) ∫∫D y(27x) dA

Now, we can evaluate these double integrals.

For x̄:

x̄ = (1/18) ∫[0,1]∫[0,√(x)] x(27x) dy dx

x̄ = (1/18) ∫[0,1] 27x^2 (√(x) - 0) dx

x̄ = (1/18) ∫[0,1] 27x^(5/2) dx

x̄ = (1/18) [18x^(7/2)/7/2] evaluated from x = 0 to x = 1

x̄ = (1/18) (7/2) (1^(7/2) - 0^(7/2))

x̄ = (7/18) (1)

x̄ = 7/18

For ȳ:

ȳ = (1/18) ∫[0,1]∫[0,√(x)] y(27x) dy dx

ȳ = (1/18) ∫[0,1] 27xy^2 dy dx

ȳ = (1/18) ∫[0,1] 27x(√(x))^2 dx

ȳ = (1/18) ∫[0,1] 27x^2 dx

ȳ = (1/18) [27x^3/3] evaluated from x = 0 to x = 1

ȳ = (1/18) (1^3/3 - 0^3/3)

ȳ = (1/18) (1/3)

ȳ = 1/54

Therefore, the center of mass of the lamina is (x̄, ȳ) = (7/18, 1/54).