Stuck on these....

[MarissaDelozier]

Really hard problems. Keep getting stuck. Here's a couple:

volume of a solid generated by revolving region:
y= x2, y= 4x - x2 (x2 means x squared)



2. arc length of graph of curve
y=x4/8 + 1/ 4x2 over interval [1,2] (x4/8 means x to the fourth over 8)

 

[skeeter]

Really hard problems. Keep getting stuck. Here's a couple:

volume of a solid generated by revolving region:
y= x2, y= 4x - x2 (x2 means x squared)

revolve this region around what? the x-axis, y-axis ???



2. arc length of graph of curve
y=x4/8 + 1/ 4x2 over interval [1,2] (x4/8 means x to the fourth over 8)

arc length = INT{a to b} sqrt[1 + (dy/dx)^2] dx

y = (x^4)/8 + 1/(4x^2)

dy/dx = (x^3)/2 - 1/(2x^3)

square dy/dx and use the integral for arc length.

 

[soroban]

Hello, Marissa!

1. Volume of a solid generated by revolving region: \(\displaystyle y\:=\:x^2,\;y\:=\:4x\,-\,x^2\)

         *        |        *
                  |
          *       |    * *(2,2)
                  | *:::::::*
            *     |*::::*    *
              *   |:::*        4
      -----------***----------*--
                  |0
                 *|            *

Revolved about what?

2. Arc length of graph of curve: \(\displaystyle \,y\:=\:\frac{x^4}{8}\,+\,\frac{1}{4x^2}\,\) over interval \(\displaystyle [1,\,2]\)

I assume you know the formula: \(\displaystyle \L\:L\;=\;\int^b_a\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2}\.dx\)

We have: \(\displaystyle \L\,y\:=\:\frac{1}{8}x^4\,+\,\frac{1}{4}x^{-2}\)

Then: \(\displaystyle \L\:\frac{dy}{dx}\:=\:\frac{1}{2}x^3\,-\,\frac{1}{2}x^{-3}\)

And: \(\displaystyle \L\;\left(\frac{dy}{dx}\right)^2\;=\;\left(\frac{1}{2}x^3 \,-\,\frac{1}{2}x^{-3}\right)^2 \;= \;\frac{1}{4}x^6\,-\,\frac{1}{2}\,+\,\frac{1}{4}x^{-6}\)

Hence: \(\displaystyle \L\;1\,+\,\left(\frac{dy}{dx}\right)^2\;=\;1\,+\,\frac{1}{4}x^6\,-\,\frac{1}{2}\,+\,\frac{1}{4}x^{-6}\)
. . . . . \(\displaystyle \L=\; \frac{1}{4}x^6\,+\,\frac{1}{2}\,+\,+\frac{1}{4}x^{-6}\;=\;\left(\frac{1}{2}x^3\,+\,\frac{1}{2}x^{-3}\right)^2\)


Therefore: \(\displaystyle \L\;\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2}\;=\;\sqrt{\left(\frac{1}{2}x^3\,+\,\frac{1}{2}x^{-3}\right)^2} \;= \;\frac{1}{2}x^3\,+\,\frac{1}{2}x^{-3}\)

Now you must evaluate: \(\displaystyle \L\:\frac{1}{2}\int^{\;\;\;2}_1\left(x^3\,+\,x^{-3}\right)\,dx\;\) . . . okay?

 

[MarissaDelozier]

question #1: about line x=2

question #2- I don't know how to evaluate the integration with the 1/2 in front. I am so confused- I can follow your steps - but cant put it all together on my own.

such as...
integrate S pie-0 xsin2x dx (S pie-0 means the curly s with pie at top and zero at bottom)
expecially confused with sin/cos in equations.

 

[skeeter]

question #1: about line x=2

one method is to use cylindrical shells ...

V = 2pi*INT{0 to 2} (2 - x)(4x - 2x^2)dx

question #2- I don't know how to evaluate the integration with the 1/2 in front. I am so confused- I can follow your steps - but cant put it all together on my own.

The 1/2 is just a constant multiplying the value of the definite integral.

The two applications of integration that you posted indicate that you should be more than familiar with the process of integration in general. Soroban has already completed the difficult part of the arc length problem.

such as...
integrate S pie-0 xsin2x dx (S pie-0 means the curly s with pie at top and zero at bottom)

do you know integration by parts? u = x, dv = sin(2x) dx
du = dx, v = -(1/2)cos(2x)

expecially confused with sin/cos in equations.

 

[galactus]

question #1: about line x=2
Does it matter which you use, shells or washers?.

Oh well, we'll start with shells. It's a little easier for this problem. For

washers we'd have to express the functions in terms of y. It can be done

though. If you can, it's good to do to see if you arrive at the same thing.

\(\displaystyle \L\\2{\pi}\int_{0}^{2}(2-x)(4x-x^{2}-(x^{2}))dx\)

\(\displaystyle \L\\2{\pi}\int_{0}^{2}(2-x)(4x-2x^{2})dx\)

What the heck, let's do the washers too:

\(\displaystyle \L\\{\pi}\int_{0}^{4}((2-(\sqrt{4-y}+2))^{2}-(2-\sqrt{y})^{2})dy\)

question #2- I don't know how to evaluate the integration with the 1/2 in front. I am so confused- I can follow your steps - but cant put it all together on my own. [
All that means is multiply it all by 1/2 when you're finished.

such as...
integrate S pie-0 xsin2x dx (S pie-0 means the curly s with pie at top and zero at bottom)
expecially confused with sin/cos in equations.

Try IBP(integration by parts)

\(\displaystyle \L\\u=x\ dv=sin(2x)\ du=dx\ v=-\frac{1}{2}cos(2x)\)

\(\displaystyle \L\\\int{xsin(2x)dx=-\frac{1}{2}xcos(2x)+\frac{1}{2}\int{cos(2x)}dx\)