To find the volume $V$ of the solid $E$ enclosed by the given surfaces, we can use a triple integral in the order $dz \, dy \, dx$. Here are the surfaces:

• $z = 4 - y^2$ (upper surface depending on $y$),
• $z = 36 - x$ (upper surface depending on $x$),
• $x = 0$ (boundary in the $x$-direction),
• $z = 0$ (lower boundary in the $z$-direction).

Step 1: Determine the bounds

1. In the $x$-direction: $x$ varies from 0 to 36 (since $z = 36 - x$ intersects $z = 0$ at $x = 36$).
2. In the $y$-direction: $y$ varies between $-2$ and $2$ (because $z = 4 - y^2$ intersects $z = 0$ when $y = \pm 2$).
3. In the $z$-direction: $z$ varies between the lower bound $z = 0$ and the smaller of $z = 4 - y^2$ and $z = 36 - x$.

The volume can be expressed as: $V = \int_{x=0}^{36} \int_{y=-2}^{2} \int_{z=0}^{\min(4-y^2, 36-x)} 1 \, dz \, dy \, dx$

Step 2: Split into regions

Since $z$ is the minimum of $4 - y^2$ and $36 - x$, we need to split the volume calculation into two regions:

• Region 1: $36 - x \leq 4 - y^2$, meaning $x \geq 32$ (from equating $36 - x = 4 - y^2$).
• Region 2: $36 - x > 4 - y^2$, meaning $x < 32$.

Thus:

1. For $0 \leq x \leq 32$, $z$ varies from $0$ to $4 - y^2$.
2. For $32 \leq x \leq 36$, $z$ varies from $0$ to $36 - x$.

Step 3: Write and compute the integrals

Region 1 ($0 \leq x \leq 32$):

$V_1 = \int_{x=0}^{32} \int_{y=-2}^{2} \int_{z=0}^{4-y^2} 1 \, dz \, dy \, dx$ The $z$-integral evaluates to $4 - y^2$, giving: $V_1 = \int_{x=0}^{32} \int_{y=-2}^{2} (4 - y^2) \, dy \, dx$ The $y$-integral is: $\int_{y=-2}^{2} (4 - y^2) \, dy = \left[ 4y - \frac{y^3}{3} \right]_{-2}^{2} = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) = \frac{64}{3}$ Thus: $V_1 = \int_{x=0}^{32} \frac{64}{3} \, dx = \frac{64}{3} \cdot 32 = \frac{2048}{3}$

Region 2 ($32 \leq x \leq 36$):

$V_2 = \int_{x=32}^{36} \int_{y=-2}^{2} \int_{z=0}^{36-x} 1 \, dz \, dy \, dx$ The $z$-integral evaluates to $36 - x$, giving: $V_2 = \int_{x=32}^{36} \int_{y=-2}^{2} (36 - x) \, dy \, dx$ The $y$-integral is: $\int_{y=-2}^{2} (36 - x) \, dy = (36 - x) \int_{y=-2}^{2} 1 \, dy = (36 - x) \cdot 4 = 4(36 - x)$ Thus: $V_2 = \int_{x=32}^{36} 4(36 - x) \, dx = 4 \int_{x=32}^{36} (36 - x) \, dx$ The integral is: $\int_{x=32}^{36} (36 - x) \, dx = \left[ 36x - \frac{x^2}{2} \right]_{32}^{36} = \left( 1296 - 648 \right) - \left( 1152 - 512 \right) = 16$ Thus: $V_2 = 4 \cdot 16 = 64$

Step 4: Total volume

$V = V_1 + V_2 = \frac{2048}{3} + 64 = \frac{2048}{3} + \frac{192}{3} = \frac{2240}{3}$

Final Answer:

$V = \frac{2240}{3} \, \text{cubic units}.$

Let me know if you need detailed calculations or visualizations!

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Related Questions:

1. What are the cross-sectional areas of the solid at specific values of $x$?
2. How do the bounds of integration change if $z = 36 - x$ were replaced with another plane?
3. What is the effect of extending the $y$-range beyond $-2$ to $2$?
4. How would the volume change if $z = 4 - y^2$ were rotated around the $z$-axis?
5. Can the solid's volume be determined using cylindrical or spherical coordinates?

Tip:

Always visualize the solid and its boundaries to clarify the limits of