find the volume of the solid obtained when the region under the curve y=xsqrt(4-x^2) from x=0 to x=2 is rotated about the y-axis.

To find the volume of the solid obtained by rotating the region under the curve \( y = x\sqrt{4 - x^2} \) from \( x = 0 \) to \( x = 2 \) around the **y-axis**, we use the method of cylindrical shells. 

The formula for the volume of a solid generated by revolving a region around the y-axis is:

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

Here, \( f(x) = x\sqrt{4 - x^2} \), and the limits of integration are \( x = 0 \) to \( x = 2 \).

Thus, the volume is given by:

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

Simplify the integrand:

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

### Step 1: Substitution
We perform the substitution:

\[
u = 4 - x^2 \quad \Rightarrow \quad du = -2x \, dx
\]

When \( x = 0 \), \( u = 4 \), and when \( x = 2 \), \( u = 0 \).

Rewriting the integral in terms of \( u \):

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

Flipping the limits and simplifying:

\[
V = 2\pi \int_{0}^{4} (4 - u)\sqrt{u} \, du
\]

### Step 2: Distribute and Integrate
Now, distribute \( \sqrt{u} \) to each term:

\[
V = 2\pi \int_{0}^{4} \left( 4\sqrt{u} - u\sqrt{u} \right) \, du
\]

This becomes:

\[
V = 2\pi \left( 4 \int_{0}^{4} u^{1/2} \, du - \int_{0}^{4} u^{3/2} \, du \right)
\]

Now, integrate each term:

\[
\int u^{1/2} \, du = \frac{2}{3} u^{3/2}, \quad \int u^{3/2} \, du = \frac{2}{5} u^{5/2}
\]

### Step 3: Evaluate the Integrals
Now substitute back the limits of integration:

\[
4 \int_{0}^{4} u^{1/2} \, du = 4 \cdot \frac{2}{3} \cdot 4^{3/2} = 4 \cdot \frac{2}{3} \cdot 8 = \frac{64}{3}
\]

\[
\int_{0}^{4} u^{3/2} \, du = \frac{2}{5} \cdot 4^{5/2} = \frac{2}{5} \cdot 32 = \frac{64}{5}
\]

### Step 4: Combine the Results
Now subtract the integrals:

\[
V = 2\pi \left( \frac{64}{3} - \frac{64}{5} \right)
\]

To combine the terms, get a common denominator:

\[
V = 2\pi \left( \frac{320}{15} - \frac{192}{15} \right) = 2\pi \cdot \frac{128}{15}
\]

Finally:

\[
V = \frac{256\pi}{15}
\]

Thus, the volume of the solid is \( \frac{256\pi}{15} \).

---

Do you want further details on the steps? Here are 5 related questions:

1. How would you calculate the volume if the region was rotated around the x-axis?
2. Can this method be used for curves that intersect the axis of rotation?
3. How does the cylindrical shell method compare to the disk method?
4. What is the geometric significance of the substitution used in this problem?
5. Can the integral limits change if the curve is shifted vertically?

**Tip**: When rotating regions, always visualize the axis of rotation to choose the right method: shells (for rotation around a vertical axis) or disks (for rotation around a horizontal axis).

Find the volume of the solid obtained when the region under the curve y = x√(4 - x^2) from x = 0 to x = 2 is rotated about the y-axis.

Learn how to calculate the volume of the solid formed by rotating the region under the curve y = x√(4 - x^2) from x = 0 to x = 2 about the y-axis using the cylindrical shell method. Detailed steps cover substitution and integration techniques, suitable for college-level Calculus II students.

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