Finding the Volume of Revolution for $ X = \sqrt{Y} $ and $ X = \frac{Y}{4} $

This problem asks us to find the volume of the solid generated when the region bounded by the curves $ X = \sqrt{Y} $ and $ X = \frac{Y}{4} $ is revolved around the x-axis.

Step 1: Express Curves in terms of $ Y $ as a function of $ X $

First, let's rewrite the given equations so that $ Y $ is expressed as a function of $ X $.

• The first curve is $ X = \sqrt{Y} $. Squaring both sides gives $ Y = X^2 $ (Note: this is valid for $ X \ge 0 $).
• The second curve is $ X = \frac{Y}{4} $. Multiplying by 4 gives $ Y = 4X $.

Step 2: Determine the Intersection Points

To find the limits of integration, we need to find where these two curves intersect. We set the expressions for $ Y $ equal to each other:

$ X^2 = 4X $

Rearrange the equation to solve for $ X $:

$ X^2 - 4X = 0 $

Factor out $ X $:

$ X(X - 4) = 0 $

This gives us two possible values for $ X $: $ X = 0 $ and $ X = 4 $.

Now, find the corresponding $ Y $ values:

• When $ X = 0 $, $ Y = 0^2 = 0 $ and $ Y = 4(0) = 0 $. The intersection point is (0, 0).
• When $ X = 4 $, $ Y = 4^2 = 16 $ and $ Y = 4(4) = 16 $. The intersection point is (4, 16).

So, the curves intersect at $ X = 0 $ and $ X = 4 $. These will be our bounds for integration.

Step 3: Identify the Outer and Inner Radii

We are revolving the region around the x-axis. We need to determine which function defines the outer radius ($ R_{outer} $) and which defines the inner radius ($ R_{inner} $) in the interval $ [0, 4] $. Let's test a value within the interval, for example, $ X = 2 $:

• For $ Y = 4X $, when $ X = 2 $, $ Y = 4(2) = 8 $.
• For $ Y = X^2 $, when $ X = 2 $, $ Y = 2^2 = 4 $.

Since $ 8 > 4 $ for $ X = 2 $, the curve $ Y = 4X $ is above $ Y = X^2 $ in the interval $ (0, 4) $. Therefore:

• Outer Radius ($ R_{outer} $) is $ Y = 4X $.
• Inner Radius ($ R_{inner} $) is $ Y = X^2 $.

Step 4: Set up the Volume Integral using the Washer Method

The volume $ V $ of the solid of revolution generated by revolving the region between $ Y = f(X) $ and $ Y = g(X) $ (where $ f(X) \ge g(X) $) around the x-axis from $ X=a $ to $ X=b $ is given by the washer method formula:

$ V = \int_{a}^{b} \pi [ (R_{outer})^2 - (R_{inner})^2 ] dx $

Substituting our functions and bounds:

$ V = \int_{0}^{4} \pi [ (4X)^2 - (X^2)^2 ] dx $

Simplify the expression inside the integral:

$ V = \pi \int_{0}^{4} [ 16X^2 - X^4 ] dx $

Step 5: Evaluate the Integral

Now, we find the antiderivative and evaluate it at the bounds:

$ V = \pi \left[ \frac{16X^3}{3} - \frac{X^5}{5} \right]_{0}^{4} $

Evaluate at the upper bound ($ X = 4 $):

$ \frac{16(4)^3}{3} - \frac{(4)^5}{5} = \frac{16(64)}{3} - \frac{1024}{5} = \frac{1024}{3} - \frac{1024}{5} $

Evaluate at the lower bound ($ X = 0 $):

$ \frac{16(0)^3}{3} - \frac{(0)^5}{5} = 0 - 0 = 0 $

Subtract the lower bound value from the upper bound value:

$ V = \pi \left[ \left( \frac{1024}{3} - \frac{1024}{5} \right) - 0 \right] $

Find a common denominator for the fractions:

$ V = \pi \left[ \frac{1024 \times 5}{15} - \frac{1024 \times 3}{15} \right] $

$ V = \pi \left[ \frac{5120}{15} - \frac{3072}{15} \right] $

$ V = \pi \left[ \frac{5120 - 3072}{15} \right] $

$ V = \pi \left[ \frac{2048}{15} \right] $

$ V = \frac{2048\pi}{15} $

Conclusion

The volume of the solid generated is $ \frac{2048\pi}{15} $ cubic units.