Question

Find the volume, V, of the solid obtained by rotating the bounded region in the first quadrant enclosed by the graphs of

Answer 1

Answer:

$\frac7{15}\pi$

Step-by-step explanation:

First of all, let's draw the two curves. First one (red) is a classic parabola, second (blue) is a similar curve with a tigher form. Imagine is made with paint so it's not high quality. The intersection between the two curves are easily found by solving a system of equations and are (0;0) and (1;1). At this point, the volume comes from the rotation of the shaded area, and can be obtained as difference from the rotation of the blue curve minus the red curve.

Remembering that the volume of the solid of revolution bound between $y=0; y=f(x); x=a; x=b$ is given by $\pi \int_a^b [f(x)]^2dx$, in our case the volume is given by the expression

$\pi [\int_0^1 (\sqrt[4]{x})^2-(x^2)^2 dx] = \pi[\int_0^1 \sqrt{x}-x^4 dx] = \pi[\frac23x^\frac32 -\frac15 x^5 ] _0^1 = \pi (\frac23 -\frac15) = \frac7{15}\pi$