Evaluate integral0^pi 4 3x 3x dx (b) Find the volume of the solid generated when the curve y = x3 is rotated completely about the x-axis between the ordinates x=0 and x=3.

Let $u = \cos 3x$.
Then, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = -3 \sin 3x$$

Rearrange to find $\sin 3x \, dx$:

$$\sin 3x \, dx = -\frac{1}{3} du$$

When $x=0$:

$$u = \cos(3 \cdot 0) = \cos(0) = 1$$

When $x=\pi$:

$$u = \cos(3 \cdot \pi) = \cos(3\pi) = -1$$

$$\int_0^{\pi} \cos^4 3x \sin 3x \, dx = \int_1^{-1} u^4 \left(-\frac{1}{3}\right) du$$ $$= -\frac{1}{3} \int_1^{-1} u^4 du$$

Swap the limits and change the sign:

$$= \frac{1}{3} \int_{-1}^{1} u^4 du$$

$$= \frac{1}{3} \left( \frac{(1)^5}{5} - \frac{(-1)^5}{5} \right)$$ $$= \frac{1}{3} \left( \frac{1}{5} - \left(-\frac{1}{5}\right) \right)$$ $$= \frac{1}{3} \left( \frac{1}{5} + \frac{1}{5} \right)$$ $$= \frac{1}{3} \left( \frac{2}{5} \right)$$ $$= \boxed{\frac{2}{15}}$$

6. (b) Find the volume of the solid generated when the curve $y = x^3$ is rotated completely about the x-axis between the ordinates $x=0$ and $x=3$.

The volume $V$ of a solid generated by rotating a curve $y=f(x)$ about the x-axis from $x=a$ to $x=b$ is given by:

$$V = \pi \int_a^b [f(x)]^2 dx$$

Given $y = x^3$, $a=0$, and $b=3$.

$$V = \pi \int_0^3 (x^3)^2 dx$$ $$V = \pi \int_0^3 x^6 dx$$

$$V = \pi \left[ \frac{x^{6+1}}{6+1} \right]_0^3$$ $$V = \pi \left[ \frac{x^7}{7} \right]_0^3$$

$$V = \pi \left( \frac{(3)^7}{7} - \frac{(0)^7}{7} \right)$$ $$V = \pi \left( \frac{2187}{7} - 0 \right)$$ $$V = \boxed{\frac{2187\pi}{7}}$$