To find the volume of the solid $R$ bounded by the surfaces $z = 4x$, $z = x^2$, and the planes $y = 0$, $y = 5$, we will set up an iterated integral. We aim to write this in the form:

$\iiint_R f(x, y, z) \, dV = \int_{A}^{B} \int_{C}^{D} \int_{E}^{F} \, dz \, dy \, dx$

Step 1: Determine the Boundaries for $z$

The solid is bounded by the surfaces $z = 4x$ and $z = x^2$ in terms of $z$:

1. Upper Bound: $z = 4x$
2. Lower Bound: $z = x^2$

So, for a fixed $x$ and $y$, $z$ ranges from $x^2$ to $4x$.

Thus, we have: $E = x^2 \quad \text{and} \quad F = 4x.$

Step 2: Determine the Boundaries for $y$

The solid is also bounded by $y = 0$ and $y = 5$, so: $C = 0 \quad \text{and} \quad D = 5.$

Step 3: Determine the Boundaries for $x$

To find the limits for $x$, we need to determine the points where $z = 4x$ intersects $z = x^2$. Setting these equations equal to each other:

$4x = x^2$ $x^2 - 4x = 0$ $x(x - 4) = 0$

This gives $x = 0$ and $x = 4$.

So, the bounds for $x$ are: $A = 0 \quad \text{and} \quad B = 4.$

Final Answer: Iterated Integral

The iterated integral to find the volume of $R$ is:

$\iiint_R f(x, y, z) \, dV = \int_0^4 \int_0^5 \int_{x^2}^{4x} \, dz \, dy \, dx.$

This integral gives the volume when $f(x, y, z) = 1$.

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Would you like further details on how to solve the integral?

Related Questions

1. How would the integral change if the plane $y = 5$ was replaced with $y = 3$?
2. How do we interpret the geometric region bounded by $z = 4x$ and $z = x^2$?
3. What would the limits of integration be if we swapped the order of integration to $dx \, dy \, dz$?
4. How can we confirm that the region defined by these bounds represents a solid?
5. What role does the plane $y = 0$ play in defining the volume of the solid?

Tip

For complex regions, sketching the boundaries can help visualize the order and range of integration.