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

[karush]

$\tiny{M2215b.08}$
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

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

 

[MarkFL]

Examine your limits of integration more closely...:)

 

[karush]

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

 

[MarkFL]

\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...

 

[karush]

this?
$x=\sqrt{\sin^{-1}\left(y\right)}$

 

[MarkFL]

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:

\(\displaystyle V=\pi\int_0^{\Large{\frac{\pi}{2}}} \sin\left(x^2\right)2x\,dx\)

 

[karush]

\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!

 

[MarkFL]

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.

 

[karush]

ok yes I think I distributed it on the practice test7

thank for your help again