M2215b.08 Find the volume of the solid (shell meithod)

In summary, the volume of the solid with $y=\sin(x^2)$, $0\le x \le \frac{\pi}{2}$, rotated about the y-axis is $\pi-\cos\left(\frac{\pi^2}{4}\right)\approx 5.5958$.
  • #1
karush
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$\tiny{M2215b.08}$
Find the volume of the solid
\begin{align*}\displaystyle y&=\sin (x^2)\\ 0&\le x \le \frac{\pi}{2}\\
\end{align*}
about the y-axis
View attachment 7629

ok this looks like a cylindrical Shell solution
So I set it up like this,,,,, hopefully

$\begin{align*}\displaystyle
V&=\int_0^12\pi x\left(\sin x^2\right)\ dx\\
\end{align*}$
 

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  • #2
Examine your limits of integration more closely...:)
 
  • #3
\begin{align*}\displaystyle V&=\int_0^{\pi/2} x\left(\sin x^2\right)\ dx\\ \end{align*}
 
  • #4
karush said:
\begin{align*}\displaystyle V&=\int_0^{\pi/2} x\left(\sin x^2\right)\ dx\\ \end{align*}

Now, reverse the change you made to the integrand...
 
  • #5
this?
$x=\sqrt{\sin^{-1}\left(y\right)}$
 
  • #6
karush said:
this?
$x=\sqrt{\sin^{-1}\left(y\right)}$

You originally had the integrand correct, but when you corrected the limits, you changed the integrand such that the integral was still incorrect overall. You want:

\(\displaystyle V=\pi\int_0^{\Large{\frac{\pi}{2}}} \sin\left(x^2\right)2x\,dx\)
 
  • #7
\begin{align*}\displaystyle
V&=\pi\int_0^{\Large{\frac{\pi}{2}}} \sin\left(x^2\right)2x\,dx\\
&= \pi\Biggr|-\cos(x^2)\Biggr|_0^{\pi/2}\\
&=\pi\Biggr|-\cos((\pi^2/4))
+\cos(0)\Biggr|\\
&=\pi[-\cos((\pi^2/4))+1]\\
&=\pi-\cos((\pi^2/4))\approx5.5958
\end{align*}

hopefully!
 
  • #8
Let's check your result using the washer method:

\(\displaystyle V=\pi\left(\int_0^{\sin\left(\frac{\pi^2}{4}\right)} \frac{\pi^2}{4}-\arcsin(y)\,dy+\int_{\sin\left(\frac{\pi^2}{4}\right)}^{1} \left(\pi-\arcsin(y)\right)-\arcsin(y)\,dy\right)\)

Letting W|A do the grunt work...

\(\displaystyle V=\pi\left(1-\cos\left(\frac{\pi^2}{4}\right)\right)\)

This is presumably what you got, although you didn't distribute $\pi$ to both terms.
 
  • #9
ok yes I think I distributed it on the practice test7

thank for your help again
 

FAQ: M2215b.08 Find the volume of the solid (shell meithod)

1. What is the shell method?

The shell method is a mathematical technique used to find the volume of a solid of revolution by integrating the cross-sectional area of the solid with respect to the axis of revolution.

2. How is the shell method different from the disk method?

The shell method involves integrating along the axis of revolution, while the disk method involves integrating perpendicular to the axis of revolution. This leads to different expressions for calculating the volume of the solid.

3. How do you find the volume using the shell method?

To find the volume using the shell method, you need to first identify the axis of revolution and the limits of integration. Then, you can use the formula V = 2π∫(radius)(height)dx to calculate the volume of the solid.

4. When is the shell method most useful?

The shell method is most useful when the solid has a curved boundary that can be expressed as a function of x. This method can also be used when the axis of revolution is not the x or y-axis.

5. Can the shell method be used for any solid of revolution?

Yes, the shell method can be used to find the volume of any solid of revolution as long as the solid can be expressed as a function of x and the axis of revolution can be identified.

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