Questions: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. y=x^3, y=sqrt(x) ; about the x-axis v=

Solution Steps

To find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^3 \) and \( y = \sqrt{x} \) about the x-axis, we use the washer method. The volume is calculated by integrating the difference of the squares of the outer radius and the inner radius over the interval where the curves intersect. First, find the points of intersection by setting \( x^3 = \sqrt{x} \). Then, set up the integral for the volume using the formula for the washer method.

To find the points of intersection of the curves \( y = x^3 \) and \( y = \sqrt{x} \), we set the equations equal to each other:

\[ x^3 = \sqrt{x} \]

Rearranging gives:

\[ x^3 - \sqrt{x} = 0 \]

Factoring out \( \sqrt{x} \):

\[ \sqrt{x}(x^{5/2} - 1) = 0 \]

This yields the solutions \( x = 0 \) and \( x = 1 \) (the real solutions). The complex solutions are not relevant for this problem.

Using the washer method, the volume \( V \) of the solid obtained by rotating the region about the x-axis is given by the integral:

\[ V = \int_{a}^{b} \left( R(x)^2 - r(x)^2 \right) \, dx \]

where \( R(x) = \sqrt{x} \) (outer radius) and \( r(x) = x^3 \) (inner radius). The limits of integration are from \( x = 0 \) to \( x = 1 \):

\[ V = \int_{0}^{1} \left( (\sqrt{x})^2 - (x^3)^2 \right) \, dx \]

This simplifies to:

\[ V = \int_{0}^{1} \left( x - x^6 \right) \, dx \]

Now we compute the integral:

\[ V = \int_{0}^{1} (x - x^6) \, dx = \left[ \frac{x^2}{2} - \frac{x^7}{7} \right]_{0}^{1} \]

Evaluating this at the limits gives:

\[ V = \left( \frac{1^2}{2} - \frac{1^7}{7} \right) - \left( \frac{0^2}{2} - \frac{0^7}{7} \right) = \frac{1}{2} - \frac{1}{7} \]

Finding a common denominator (14):

\[ V = \frac{7}{14} - \frac{2}{14} = \frac{5}{14} \]

Final Answer