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