To solve the problem of finding the volume of the solid obtained by rotating the curve $y = x \sqrt{4 - x^2}$ from $x = 0$ to $x = 2$ about the y-axis, we will use the method of cylindrical shells. The formula for the volume of a solid generated by revolving a region about the y-axis using cylindrical shells is:

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

Here:

• $f(x) = x \sqrt{4 - x^2}$
• $a = 0$
• $b = 2$

Step 1: Set up the integral

We substitute $f(x)$ into the formula for the volume:

$V = 2 \pi \int_{0}^{2} x \cdot x \sqrt{4 - x^2} \, dx = 2 \pi \int_{0}^{2} x^2 \sqrt{4 - x^2} \, dx$

Step 2: Simplify the integral using substitution

Let $u = 4 - x^2$, then:

$du = -2x \, dx$

When $x = 0$, $u = 4$, and when $x = 2$, $u = 0$.

Thus, the integral becomes:

$V = 2 \pi \int_{4}^{0} -\frac{1}{2} (4 - u) \sqrt{u} \, du = \pi \int_{0}^{4} (4 - u) \sqrt{u} \, du$

Step 3: Split the integral

Now we can split the integral into two simpler parts:

$V = \pi \left( \int_{0}^{4} 4 \sqrt{u} \, du - \int_{0}^{4} u \sqrt{u} \, du \right)$

Step 4: Solve the integrals

1. For $\int_{0}^{4} 4 \sqrt{u} \, du$:

$\int \sqrt{u} \, du = \frac{2}{3} u^{3/2}$ $4 \int_{0}^{4} \sqrt{u} \, du = 4 \left[ \frac{2}{3} u^{3/2} \right]_{0}^{4} = 4 \times \frac{2}{3} \times 8 = \frac{64}{3}$

2. For $\int_{0}^{4} u \sqrt{u} \, du = \int_{0}^{4} u^{3/2} \, du$:

$\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$ $\int_{0}^{4} u^{3/2} \, du = \left[ \frac{2}{5} u^{5/2} \right]_{0}^{4} = \frac{2}{5} \times 32 = \frac{64}{5}$

Step 5: Combine the results

Now subtract the second integral from the first:

$V = \pi \left( \frac{64}{3} - \frac{64}{5} \right) = \pi \times \frac{320 - 192}{15} = \pi \times \frac{128}{15}$

Thus, the volume is:

$V = \frac{128 \pi}{15}$

Final Answer:

$V = \frac{128 \pi}{15} \, \text{cubic units}$

Let me know if you need further clarifications or explanations!

Here are 5 related questions:

1. What is the formula for finding the volume of solids using the disk method?
2. How do you derive the method of cylindrical shells?
3. How would the problem change if the curve was rotated about the x-axis?
4. What if the limits of integration were from $x = -2$ to $x = 2$?
5. Can you use the washer method in this problem instead of cylindrical shells?

Tip: Always check if rotating around the y-axis or x-axis to determine the best integration method!