Questions: Sketch the region bounded by the curves y=5 sqrt(x) and y=5 x^3 then find the volume of the solid generated by revolving this region about the y-axis. - 2 pi 6 pi 3 pi 4 pi 5 pi None of the above.

Solution Steps

To find the region bounded by the curves $y = 5\sqrt{x}$ and $y = 5x^3$, we first need to find their points of intersection. Set the equations equal to each other:

$$5\sqrt{x} = 5x^3$$

Divide both sides by 5:

$$\sqrt{x} = x^3$$

Square both sides to eliminate the square root:

$$x = x^6$$

Rearrange the equation:

$$x^6 - x = 0$$

Factor the equation:

$$x(x^5 - 1) = 0$$

This gives us the solutions:

$$x = 0 \quad \text{or} \quad x^5 = 1$$

Solving $x^5 = 1$ gives:

$$x = 1$$

Thus, the curves intersect at $x = 0$ and $x = 1$.

The volume of the solid generated by revolving the region about the $y$-axis can be found using the method of cylindrical shells. The formula for the volume is:

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

where $f(x) = 5\sqrt{x}$ and $g(x) = 5x^3$, and the limits of integration are from $x = 0$ to $x = 1$.

Substitute the functions and limits into the integral:

$$V = 2\pi \int_{0}^{1} x(5\sqrt{x} - 5x^3) \, dx$$

Simplify the integrand:

$$= 2\pi \int_{0}^{1} (5x^{3/2} - 5x^4) \, dx$$

Integrate term by term:

$$= 2\pi \left[ \frac{5}{\frac{5}{2}}x^{5/2} - \frac{5}{5}x^5 \right]_{0}^{1}$$

$$= 2\pi \left[ 2x^{5/2} - x^5 \right]_{0}^{1}$$

Evaluate the definite integral:

$$= 2\pi \left[ (2 \cdot 1^{5/2} - 1^5) - (2 \cdot 0^{5/2} - 0^5) \right]$$

$$= 2\pi (2 - 1)$$

$$= 2\pi$$

Final Answer