The region under the curves #y=cosxsqrtsinx, 0

Question

The region under the curves #y=cosxsqrtsinx, 0<=x<=pi/2# is rotated about the x axis.  How do you sketch the region and find the volumes of the two solids of revolution?

Charles Arocha · Accepted Answer

#V=piint_0^(pi/2)(cosxsqrtsinx)^2dx=pi/3#

A sketch is provided:

graph{cos(x)sqrt(sin(x)) [-1.5, 1.5, -1, 4]}

As #sinx>=0# for #0<=x<=pi#, #y# is defined for the interval of interest, #0<=x<=pi/2#. As such, there will only be one volume of revolution (not two, as the question implies).

When this solid is rotated around the #x#-axis, we essentially are creating an infinite amount of cylinders with a radius given by the function value at each point #0<=x<=pi/2#.

The volume will be the sum of the volumes of these infinitesimally small cylinders, whose volumes are given individually by #piy^2dx#, where #piy^2# is the area of the face of the circle and #dx# is the infinitesimally small thickness of each cylinder.

So, the volume of the entire solid will be given by the integral of this function within the bounds, or:

#int_0^(pi/2)piy^2dx=piint_0^(pi/2)(cosxsqrtsinx)^2dx=piint_0^(pi/2)cos^2xsinxdx#

Let #u=cosx#, implying that #du=-sinxdx#. This also will cause the bounds to change from #x=0=>u=cos(0)=1# and #x=pi/2=>u=cos(pi/2)=0#:

#=-piint_0^(pi/2)cos^2x(-sinxdx)=-piint_1^0u^2du=-pi[u^3/3]_1^0#

#=-pi(0^3/3-1^3/3)=-pi(-1/3)=pi/3#

Evelyn Agard · Answer

undefined To sketch the region under the curve $y = \cos(x)\sqrt{\sin(x)}$ for $0 \leq x \leq \frac{\pi}{2}$, you first need to analyze the behavior of $\cos(x)$ and $\sqrt{\sin(x)}$ separately in this interval.

1. For $\cos(x)$, as $x$ varies from $0$ to $\frac{\pi}{2}$, $\cos(x)$ starts at $1$ and decreases to $0$.
2. For $\sqrt{\sin(x)}$, as $x$ varies from $0$ to $\frac{\pi}{2}$, $\sqrt{\sin(x)}$ starts at $0$ and increases to $1$.

So, the region under the curve will be bounded by the x-axis and the curve itself.

To find the volumes of the two solids of revolution, you'll need to divide the region into two parts.

1. For the first part, rotate the region under the curve about the x-axis. This forms a solid with a "washer" shape, as there's a hole in the middle.
2. For the second part, rotate the region under the curve about the x-axis up to the curve itself. This forms a solid with a cylindrical shape.

To find the volume of each solid, you'll need to integrate using the disk (washer) or cylindrical shell method, depending on the part. For the first part, the washer method involves subtracting the volume of the inner circle from the volume of the outer circle. For the second part, you'll use cylindrical shells.

Once you set up the integrals, integrate them with respect to $x$ from $0$ to $\frac{\pi}{2}$, and evaluate to find the volumes of the two solids of revolution.

The region under the curves #y=cosxsqrtsinx, 0&lt;=x&lt;=pi/2# is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?

The region under the curves #y=cosxsqrtsinx, 0<=x<=pi/2# is rotated about the x axis. How do you sketch the region and find the volumes of the two solids of revolution?