The problem asks us to find the moments of inertia $I_x$, $I_y$, and $I_z$ for the lamina bounded by the curves $y = x^2$ and $y = 4$, with density function $\delta(x, y) = y$.

1. Problem Description

The problem asks us to find the moments of inertia $I_x$, $I_y$, and $I_z$ for the lamina bounded by the curves $y = x^2$ and $y = 4$, with density function $\delta(x, y) = y$.

2. Solution Steps

First, we need to find the intersection points of $y = x^2$ and $y = 4$.

$x^2 = 4 \implies x = \pm 2$. Thus, the region is bounded by $x=-2$ to $x=2$ and $x^2$ to $4$.

Next, we calculate the mass $m$:

$m = \int_{-2}^{2} \int_{x^2}^{4} \delta(x, y) dy dx = \int_{-2}^{2} \int_{x^2}^{4} y dy dx = \int_{-2}^{2} [\frac{1}{2}y^2]_{x^2}^{4} dx = \int_{-2}^{2} (\frac{1}{2}(16 - x^4)) dx = \int_{-2}^{2} (8 - \frac{1}{2}x^4) dx = [8x - \frac{1}{10}x^5]_{-2}^{2} = (16 - \frac{32}{10}) - (-16 + \frac{32}{10}) = 32 - \frac{64}{10} = 32 - \frac{32}{5} = \frac{160 - 32}{5} = \frac{128}{5}$.

Now, we calculate $I_x$:

$I_x = \int_{-2}^{2} \int_{x^2}^{4} y^2 \delta(x, y) dy dx = \int_{-2}^{2} \int_{x^2}^{4} y^3 dy dx = \int_{-2}^{2} [\frac{1}{4}y^4]_{x^2}^{4} dx = \int_{-2}^{2} (\frac{1}{4}(256 - x^8)) dx = \int_{-2}^{2} (64 - \frac{1}{4}x^8) dx = [64x - \frac{1}{36}x^9]_{-2}^{2} = (128 - \frac{512}{36}) - (-128 + \frac{512}{36}) = 256 - \frac{1024}{36} = 256 - \frac{256}{9} = \frac{2304 - 256}{9} = \frac{2048}{9}$.

Next, we calculate $I_y$:

$I_y = \int_{-2}^{2} \int_{x^2}^{4} x^2 \delta(x, y) dy dx = \int_{-2}^{2} \int_{x^2}^{4} x^2 y dy dx = \int_{-2}^{2} x^2 [\frac{1}{2}y^2]_{x^2}^{4} dx = \int_{-2}^{2} x^2 (\frac{1}{2}(16 - x^4)) dx = \int_{-2}^{2} (8x^2 - \frac{1}{2}x^6) dx = [\frac{8}{3}x^3 - \frac{1}{14}x^7]_{-2}^{2} = (\frac{64}{3} - \frac{128}{14}) - (-\frac{64}{3} + \frac{128}{14}) = \frac{128}{3} - \frac{256}{14} = \frac{128}{3} - \frac{128}{7} = 128(\frac{1}{3} - \frac{1}{7}) = 128(\frac{7 - 3}{21}) = 128(\frac{4}{21}) = \frac{512}{21}$.

Finally, we calculate $I_z$:

$I_z = I_x + I_y = \frac{2048}{9} + \frac{512}{21} = \frac{2048 \cdot 7 + 512 \cdot 3}{63} = \frac{14336 + 1536}{63} = \frac{15872}{63}$.

3. Final Answer

$I_x = \frac{2048}{9}$

$I_y = \frac{512}{21}$

$I_z = \frac{15872}{63}$