Calculate the integral: integral from 0 to R x²

Calculate the integral: $\int_0^R x^2 \sqrt{R^2-x^2} \, dx$.

We will use trigonometric substitution to evaluate this integral.

Step 1: Use trigonometric substitution

Let $x = R\sin\theta$. Then $dx = R\cos\theta \, d\theta$, and $\sqrt{R^2-x^2} = R\cos\theta$ for $\theta \in [0, \frac{\pi}{2}]$.

Step 2: Change the limits of integration

When $x = 0$: $\theta = 0$. When $x = R$: $\theta = \frac{\pi}{2}$.

Step 3: Rewrite the integral

\[\int_0^R x^2 \sqrt{R^2-x^2} \, dx = \int_0^{\frac{\pi}{2}} (R\sin\theta)^2 \cdot R\cos\theta \cdot R\cos\theta \, d\theta = R^4 \int_0^{\frac{\pi}{2}} \sin^2\theta \cos^2\theta \, d\theta\]

Step 4: Simplify using trigonometric identities

Using $\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$:

\[\sin^2\theta\cos^2\theta = \frac{1}{4}\sin^2(2\theta)\]

Therefore:

\[R^4 \int_0^{\frac{\pi}{2}} \sin^2\theta \cos^2\theta \, d\theta = \frac{R^4}{4} \int_0^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta\]

Step 5: Use the half-angle identity again

\[\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2}\]

Therefore:

\[\frac{R^4}{4} \int_0^{\frac{\pi}{2}} \sin^2(2\theta) \, d\theta = \frac{R^4}{8} \int_0^{\frac{\pi}{2}} (1 - \cos(4\theta)) \, d\theta\]

Step 6: Integrate

\[= \frac{R^4}{8} \left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\frac{\pi}{2}}\]

Step 7: Evaluate at the limits

At $\theta = \frac{\pi}{2}$: $\frac{\pi}{2} - \frac{\sin(2\pi)}{4} = \frac{\pi}{2}$. At $\theta = 0$: $0$.

Step 8: Calculate the final result

\[\int_0^R x^2 \sqrt{R^2-x^2} \, dx = \frac{R^4}{8} \cdot \frac{\pi}{2} = \frac{\pi R^4}{16}\]

Geometric interpretation:

This integral has a beautiful geometric meaning related to moments of inertia and centroid calculations.

• The integrand $x^2\sqrt{R^2-x^2}$ represents a weighted version of the quarter circle, where each point is weighted by the square of its distance from the $y$-axis.
• This type of integral appears in physics when calculating moments of inertia of circular regions.
• The result $\frac{\pi R^4}{16}$ can be compared to the area of the quarter circle from Problem 34: $\frac{\pi R^2}{4}$.
• Note that $\frac{\pi R^4}{16} = \frac{R^2}{4} \cdot \frac{\pi R^2}{4}$, showing a relationship between this weighted integral and the basic area.

Final answer:

\[\boxed{\int_0^R x^2 \sqrt{R^2-x^2} \, dx = \frac{\pi R^4}{16}}\]