A thin plate covers the triangular region bounded by the x-axis and the lines x = 1 and y = 2x in the first quadrant. The plate's density at the point (x, y) is δ(x, y) = 6x + 6. F... A thin plate covers the triangular region bounded by the x-axis and the lines x = 1 and y = 2x in the first quadrant. The plate's density at the point (x, y) is δ(x, y) = 6x + 6. Find the plate's mass, first moments, and center of mass about the coordinate axes. Read more

Steps to Solve

1. Determine the region of integration

The triangular region is bounded by the x-axis, the line $x = 1$, and the line $y = 2x$. The vertices of the triangle are $(0, 0)$, $(1, 0)$, and $(1, 2)$.

2. Set up the double integral for mass

The mass $M$ of the plate can be calculated using the density function, $\delta(x, y) = 6y + 6$. The mass is given by:

$$ M = \int_0^1 \int_0^{2x} (6y + 6) , dy , dx $$

3. Evaluate the inner integral

First, calculate the integral with respect to $y$:

$$ \int_0^{2x} (6y + 6) , dy = \left[ 3y^2 + 6y \right]_0^{2x} $$

Evaluate at the bounds:

$$ = 3(2x)^2 + 6(2x) = 12x^2 + 12x $$

4. Set up the outer integral

Then, substitute this result into the outer integral:

$$ M = \int_0^1 (12x^2 + 12x) , dx $$

5. Calculate the outer integral

Now evaluate:

$$ M = \left[ 4x^3 + 6x^2 \right]_0^1 = 4(1)^3 + 6(1)^2 = 4 + 6 = 10 $$

Thus, the mass of the plate is:

$$ M = 10 $$

6. Find the first moments about the axes

The first moment about the x-axis ($M_y$) is given by:

$$ M_y = \int_0^1 \int_0^{2x} y (6y + 6) , dy , dx $$

7. Evaluate the inner integral for $M_y$

Calculating the inner integral:

$$ \int_0^{2x} y (6y + 6) , dy = \int_0^{2x} (6y^2 + 6y) , dy = \left[ 2y^3 + 3y^2 \right]_0^{2x} $$

Evaluate:

$$ = 2(2x)^3 + 3(2x)^2 = 16x^3 + 12x^2 $$

8. Set up the outer integral for $M_y$

Now substitute into the outer integral:

$$ M_y = \int_0^1 (16x^3 + 12x^2) , dx $$

9. Calculate the outer integral for $M_y$

$$ M_y = \left[ 4x^4 + 4x^3 \right]_0^1 = 4(1)^4 + 4(1)^3 = 4 + 4 = 8 $$

10. Find the first moment about the y-axis ($M_x$)

The first moment about the y-axis is calculated similarly:

$$ M_x = \int_0^1 \int_0^{2x} x (6y + 6) , dy , dx $$

11. Evaluate the inner integral for $M_x$

Calculating the inner integral:

$$ \int_0^{2x} x (6y + 6) , dy = x \int_0^{2x} (6y + 6) , dy $$

We already found that:

$$ \int_0^{2x} (6y + 6) , dy = 12x^2 + 12x $$

Thus,

$$ M_x = \int_0^1 x(12x^2 + 12x) , dx = \int_0^1 (12x^3 + 12x^2) , dx $$

12. Calculate the outer integral for $M_x$

$$ M_x = \left[ 3x^4 + 4x^3 \right]_0^1 = 3 + 4 = 7 $$

13. Find the center of mass

The coordinates of the center of mass ($\bar{x}, \bar{y}$) are given by:

$$ \bar{x} = \frac{M_y}{M} = \frac{8}{10} = 0.8 $$

$$ \bar{y} = \frac{M_x}{M} = \frac{7}{10} = 0.7 $$