Question #6a05f

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Question #6a05f

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Answer 1 · 2017-04-03T23:56:24

The volume is $π^{2} - 2 π$ $pi^2-2pi$ cubic units

The integral using the method of disks is

$π \int_{0}^{1} {(arccos (x))}^{2} d y$ $piint_0^1(arccos(x))^2dy$

Explanation:

From the figure I am assuming that the problem is rotating $cos (x)$ $cos(x)$ (in the first quadrant) from $x = 0$ $x=0$ to $x = \frac{π}{2}$ $x=pi/2$ about the $y$ $y$ axis.

Using the shell method our radius will be some value of $x$ $x$ over the interval $[0, \frac{π}{2}]$ $[0,pi/2]$

The shell height is just the function $y = cos (x)$ $y=cos(x)$

So the integral using the shell method is

$2 π \int_{0}^{\frac{π}{2}} x cos (x) d x$ $2piint_0^(pi/2)xcos(x)dx$

We can evaluate this integral using integration by parts.

Let $u = x$ $u=x$ then $d u = d x$ $du=dx$

Let $d v = cos (x) d x$ $dv=cos(x)dx$ the $\int d v = \int cos (x) d x$ $intdv=intcos(x)dx$
so $v = sin (x)$ $v=sin(x)$

The parts formula is $u v - \int v d u$ $uv-intvdu$

Proceeding we have

$2 π [x sin (x) - \int sin (x) d x]$ $2pi[xsin(x)-intsin(x)dx]$

$2 π [x sin (x) - (- cos (x))]$ $2pi[xsin(x)-(-cos(x))]$

$2 π [x sin (x) + cos (x)]$ $2pi[xsin(x)+cos(x)]$

Evaluating

$2 π [(\frac{π}{2}) sin ((\frac{π}{2})) + cos ((\frac{π}{2}))) - (0 + 1)]$ $2pi[(pi/2)sin((pi/2))+cos((pi/2)))-(0+1)]$

$2 π [(\frac{π}{2}) - 1]$ $2pi[(pi/2)-1]$

$π^{2} - 2 π$ $pi^2-2pi$

Now onward with the disk method.
We have to get things in terms of $y$ $y$

So $x = arccos (y)$ $x=arccos(y)$

We will integrate over the interval $y = 0$ $y=0$ to $y = 1$ $y=1$

Our radius is $x = arccos (x)$ $x=arccos(x)$

Using the method of disks the integral for the volume is

$π \int_{0}^{1} {(arccos (y))}^{2} d y$ $piint_0^1(arccos(y))^2dy$

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Question #6a05f

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