We are asked to find the moments of inertia $I_x$, $I_y$, and $I_z$ for the lamina bounded by the given curves and with the indicated density $\delta$ for problem 11. The curves are $y = \sqrt{x}$, $x = 9$, and $y = 0$. The density is $\delta(x, y) = x + y$.

1. Problem Description

We are asked to find the moments of inertia $I_x$, $I_y$, and $I_z$ for the lamina bounded by the given curves and with the indicated density $\delta$ for problem

1

1. The curves are $y = \sqrt{x}$, $x = 9$, and $y = 0$. The density is $\delta(x, y) = x + y$.

2. Solution Steps

First, we need to determine the limits of integration. The region is bounded by $y = \sqrt{x}$, $x = 9$, and $y = 0$. This means $0 \le y \le \sqrt{x}$ and $0 \le x \le 9$. We can also describe the region as $0 \le x \le 9$ and $0 \le y \le \sqrt{x}$.

The formulas for the moments of inertia are:

$I_x = \iint y^2 \delta(x, y) \, dA$

$I_y = \iint x^2 \delta(x, y) \, dA$

$I_z = \iint (x^2 + y^2) \delta(x, y) \, dA$

We have $\delta(x, y) = x + y$.

Now, we calculate $I_x$:

$I_x = \int_0^9 \int_0^{\sqrt{x}} y^2 (x + y) \, dy \, dx = \int_0^9 \int_0^{\sqrt{x}} (xy^2 + y^3) \, dy \, dx$

$I_x = \int_0^9 \left[ \frac{xy^3}{3} + \frac{y^4}{4} \right]_0^{\sqrt{x}} dx = \int_0^9 \left( \frac{x(x^{3/2})}{3} + \frac{x^2}{4} \right) dx = \int_0^9 \left( \frac{x^{5/2}}{3} + \frac{x^2}{4} \right) dx$

$I_x = \left[ \frac{2x^{7/2}}{21} + \frac{x^3}{12} \right]_0^9 = \frac{2(9)^{7/2}}{21} + \frac{9^3}{12} = \frac{2(3^2)^{7/2}}{21} + \frac{729}{12} = \frac{2(3^7)}{21} + \frac{243}{4} = \frac{2(2187)}{21} + \frac{243}{4} = \frac{4374}{21} + \frac{243}{4} = \frac{1458}{7} + \frac{243}{4} = \frac{4(1458) + 7(243)}{28} = \frac{5832 + 1701}{28} = \frac{7533}{28}$

Now, we calculate $I_y$:

$I_y = \int_0^9 \int_0^{\sqrt{x}} x^2 (x + y) \, dy \, dx = \int_0^9 \int_0^{\sqrt{x}} (x^3 + x^2y) \, dy \, dx$

$I_y = \int_0^9 \left[ x^3y + \frac{x^2y^2}{2} \right]_0^{\sqrt{x}} dx = \int_0^9 \left( x^3\sqrt{x} + \frac{x^2(x)}{2} \right) dx = \int_0^9 \left( x^{7/2} + \frac{x^3}{2} \right) dx$

$I_y = \left[ \frac{2x^{9/2}}{9} + \frac{x^4}{8} \right]_0^9 = \frac{2(9)^{9/2}}{9} + \frac{9^4}{8} = \frac{2(3^2)^{9/2}}{9} + \frac{6561}{8} = \frac{2(3^9)}{9} + \frac{6561}{8} = \frac{2(19683)}{9} + \frac{6561}{8} = \frac{39366}{9} + \frac{6561}{8} = \frac{13122}{3} + \frac{6561}{8} = \frac{8(13122) + 3(6561)}{24} = \frac{104976 + 19683}{24} = \frac{124659}{24} = \frac{41553}{8}$

Now, we calculate $I_z$:

$I_z = \int_0^9 \int_0^{\sqrt{x}} (x^2 + y^2) (x + y) \, dy \, dx = \int_0^9 \int_0^{\sqrt{x}} (x^3 + x^2y + xy^2 + y^3) \, dy \, dx$

$I_z = \int_0^9 \left[ x^3y + \frac{x^2y^2}{2} + \frac{xy^3}{3} + \frac{y^4}{4} \right]_0^{\sqrt{x}} dx = \int_0^9 \left( x^3\sqrt{x} + \frac{x^2(x)}{2} + \frac{x(x^{3/2})}{3} + \frac{x^2}{4} \right) dx = \int_0^9 \left( x^{7/2} + \frac{x^3}{2} + \frac{x^{5/2}}{3} + \frac{x^2}{4} \right) dx$

$I_z = \left[ \frac{2x^{9/2}}{9} + \frac{x^4}{8} + \frac{2x^{7/2}}{21} + \frac{x^3}{12} \right]_0^9 = \frac{2(9)^{9/2}}{9} + \frac{9^4}{8} + \frac{2(9)^{7/2}}{21} + \frac{9^3}{12} = \frac{2(3^9)}{9} + \frac{6561}{8} + \frac{2(3^7)}{21} + \frac{729}{12} = \frac{39366}{9} + \frac{6561}{8} + \frac{4374}{21} + \frac{243}{4} = \frac{13122}{3} + \frac{6561}{8} + \frac{1458}{7} + \frac{243}{4} = \frac{73584+19683+4914+1701}{24} = \frac{104976 + 6561 \times 3 + 1458 \times \frac{24}{7} + 243 \times 6}{24}$

$= \frac{13122}{3} + \frac{6561}{8} + \frac{1458}{7} + \frac{243}{4} = \frac{36672}{8} + \frac{6561}{8} + \frac{5832}{21} + \frac{1701}{28}$

$I_z = I_x + I_y = \frac{7533}{28} + \frac{41553}{8} = \frac{2(7533) + 7(41553)}{56} = \frac{15066 + 290871}{56} = \frac{305937}{56}$

3. Final Answer

$I_x = \frac{7533}{28}$

$I_y = \frac{41553}{8}$

$I_z = \frac{305937}{56}$