Find the volume of the solid obtained by rotating the region enclosed by the graphs y = x^2, y = x^(1/2) about the line x = -3. (Use symbolic notation and fractions where needed.)

Steps to Solve

1. Find Points of Intersection

To determine the enclosed area, find where the curves intersect by setting them equal to each other:

$$ x^2 = x^{1/2} $$

Solve for $x$:

$$ x^2 - x^{1/2} = 0 $$

Factoring gives:

$$ x^{1/2}(x^{3/2} - 1) = 0 $$

This leads to solutions $x = 0$ and $x = 1$.

2. Sketch the Region

Identify the region bounded by the curves $y = x^2$ and $y = x^{1/2}$ between $x = 0$ and $x = 1$. The top curve is $y = x^{1/2}$ and the bottom curve is $y = x^2$.

3. Set Up the Volume Integral

Since we are rotating around the line $x = -3$, we'll use the shell method. The formula for volume using shell method is:

$$ V = 2\pi \int_{a}^{b} (radius)(height),dx $$

Here, the radius is $|x + 3|$ and the height is $x^{1/2} - x^2$:

$$ V = 2\pi \int_0^1 (x + 3)(x^{1/2} - x^2) ,dx $$

4. Integrate the Function

Now we need to compute the integral:

$$ V = 2\pi \int_0^1 (x + 3)(x^{1/2} - x^2) ,dx $$

Expand the integrand:

$$ (x + 3)(x^{1/2} - x^2) = x^{3/2} - x^3 + 3x^{1/2} - 3x^2 $$

Therefore,

$$ V = 2\pi \int_0^1 (x^{3/2} - x^3 + 3x^{1/2} - 3x^2) ,dx $$

5. Compute Each Integral

Now, compute the integral term by term:

$$ \int x^{3/2} ,dx = \frac{2}{5} x^{5/2} $$

$$ \int x^3 ,dx = \frac{1}{4} x^4 $$

$$ \int x^{1/2} ,dx = \frac{2}{3} x^{3/2} $$

$$ \int x^2 ,dx = \frac{1}{3} x^3 $$

Evaluating the definite integrals from 0 to 1 gives:

$$ \left[ \frac{2}{5}(1)^{5/2} - \frac{1}{4}(1)^4 + 3\left(\frac{2}{3}(1)^{3/2}\right) - 3\left(\frac{1}{3}(1)^3\right) \right] $$

Which simplifies to:

$$ \frac{2}{5} - \frac{1}{4} + 2 - 1 = \frac{2}{5} - \frac{1}{4} + 1 $$

6. Calculate Volume

Combine the results and multiply by $2\pi$:

$$ V = 2\pi \left( \frac{2}{5} - \frac{1}{4} + 1 \right) $$

Convert to a common denominator to simplify:

Common denominator for 5, 4 is 20, hence:

$$ V = 2\pi \left( \frac{8}{20} - \frac{5}{20} + \frac{20}{20} \right) $$

$$ V = 2\pi \left( \frac{23}{20} \right) $$

Finally, compute:

$$ V = \frac{46\pi}{20} = \frac{23\pi}{10} $$