The problem asks for the volume of the solid formed by revolving the region enclosed by the curves $y = 3\sqrt{4x}$ and $y = 7 - x$ about the x-axis.

We will use the method of disks (or washers) to calculate the volume of the solid. The formula for the volume of a solid obtained by rotating a region around the x-axis is:

$V = \pi \int_{a}^{b} \left[ (f(x))^2 - (g(x))^2 \right] \, dx$

Where:

• $f(x)$ is the function describing the outer radius (in this case, the line $y = 7 - x$).
• $g(x)$ is the function describing the inner radius (in this case, the curve $y = 3\sqrt{4x}$).
• The limits of integration $a$ and $b$ correspond to the x-values where the two curves intersect.

Step 1: Find the points of intersection

To find the limits of integration, set the two equations equal to each other:

$3\sqrt{4x} = 7 - x$

First, square both sides to eliminate the square root:

$(3\sqrt{4x})^2 = (7 - x)^2$ $9(4x) = (7 - x)^2$ $36x = 49 - 14x + x^2$

Rearrange the terms to set the equation equal to 0:

$x^2 - 50x + 49 = 0$

Now solve this quadratic equation using the quadratic formula:

$x = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(49)}}{2(1)}$ $x = \frac{50 \pm \sqrt{2500 - 196}}{2}$ $x = \frac{50 \pm \sqrt{2304}}{2}$ $x = \frac{50 \pm 48}{2}$

Thus, the two solutions are:

$x = \frac{50 + 48}{2} = 49 \quad \text{and} \quad x = \frac{50 - 48}{2} = 1$

So the curves intersect at $x = 1$ and $x = 49$.

Step 2: Set up the integral for the volume

We integrate from $x = 0$ to $x = 1$ (where the region is bounded by the curves):

$V = \pi \int_0^1 \left[ (7 - x)^2 - (3\sqrt{4x})^2 \right] \, dx$

Step 3: Simplify the integrand

We have:

$(7 - x)^2 = 49 - 14x + x^2$

$(3\sqrt{4x})^2 = 9(4x) = 36x$

Thus, the volume integral becomes:

$V = \pi \int_0^1 \left[ (49 - 14x + x^2) - 36x \right] \, dx$ $V = \pi \int_0^1 \left[ 49 - 50x + x^2 \right] \, dx$

Step 4: Evaluate the integral

Now integrate term-by-term:

$\int_0^1 49 \, dx = 49x \bigg|_0^1 = 49$ $\int_0^1 -50x \, dx = -50 \times \frac{x^2}{2} \bigg|_0^1 = -50 \times \frac{1}{2} = -25$ $\int_0^1 x^2 \, dx = \frac{x^3}{3} \bigg|_0^1 = \frac{1}{3}$

Thus, the volume is:

$V = \pi \left( 49 - 25 + \frac{1}{3} \right)$ $V = \pi \left( 24 + \frac{1}{3} \right)$ $V = \pi \times \frac{73}{3}$

So, the volume of the solid is:

$V = \frac{73\pi}{3}$

This is the exact volume of the solid obtained by revolving the region about the x-axis.

If you have any further questions or need more details, feel free to ask!

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Further Questions:

1. How can the volume formula change if the region is rotated around a different axis, such as the y-axis?
2. What are some other methods to find the volume of solids of revolution?
3. How can you calculate the volume of a solid using cylindrical shells instead of disks?
4. What is the impact of changing the limits of integration in such volume problems?
5. Can the volume of the solid be determined without using calculus, and if so, how?

Tip:

For complex volume problems, always carefully determine the points of intersection first, as they define the bounds for your integral.