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Integration application: solids of revolution

paulmdrdo

Active member
May 13, 2013
386
need some help here.

1. find the volume (by washer method) of the solid generated by revolving the region bounded by $y=3+x^2$ and the line $y=4$ about the x-axis.

2. write the integral that will give the volume of the solid generated by revolving the region bounded by $y=e^{2x}$, $x=-1$ and $x=2$ about x=-2. (by cylindrical shell method).

thanks!
 
Last edited:

Petrus

Well-known member
Feb 21, 2013
739
Re: Integration application.

need some help here.

1. find the volume (by washer method) of the solid generated by revolving the region bounded by $y=3+x^2$ and the line $y=4$ about the x-axis.

2. write the integral that will give the volume of the solid generated by revolving the region by $y=e^{2x}$, $x=1$ and $x=2$. (by cylindrical shell method).

thanks!
Hello,
1.Do you got any progress? Do you know how to solve the integral region ( the a and b in \(\displaystyle \int_a^b\)

solve \(\displaystyle 3+x^2=4\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Integration application.

I recommend you do a search here on the word "revolution" and you will find many topics that demonstrate how to find the volumes of solids of revolution.

I generally recommend that the region to be rotated and the axis of rotation first be sketched. This will help you greatly in the next step...

Then determine the volume of an element of the solid, whether it be a disk, washer or shell.

Finally, use integration to sum all of the elements to get the total volume. As Petrus points out, you will need to determine the limits of integration.

After looking at some of our other topics here in your search, and then attempting the steps I outlined above, show us what you find, and we can then help you from there if you get stuck.

By the way, you did not state the axis of rotation for the second problem. I would assume it is a vertical axis.
 

paulmdrdo

Active member
May 13, 2013
386
the limits of integration in 1 is x=1 and x=-1.

$\displaystyle\int_{-1}^1 \pi(7-6x-x^4)dx$

my answer is $\displaystyle \frac{68\pi}{5}cu.\,units$

help me in number 2 please

2. write the integral that will give the volume of the solid generated by revolving the region bounded by $y=e^{2x}$, $x=-1$ and $x=2$ about x=-2. (by cylindrical shell method).
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The first one is incorrect. You want:

\(\displaystyle V=2\pi\int_0^1 4^2-\left(3+x^2 \right)^2\,dx\)

I am using the even function rule to make evaluating the definite integral easier.

Your simplification (expansion and collection of like terms in the integrand) is almost correct, and I suspect it was simply a careless mistake rather than a mistake in actual technique.

For the second one, have you drawn a diagram and determined the volume of an arbitrary shell?
 

paulmdrdo

Active member
May 13, 2013
386
The first one is incorrect. You want:

\(\displaystyle V=2\pi\int_0^1 4^2-\left(3+x^2 \right)^2\,dx\)

I am using the even function rule to make evaluating the definite integral easier.

Your simplification (expansion and collection of like terms in the integrand) is almost correct, and I suspect it was simply a careless mistake rather than a mistake in actual technique.

For the second one, have you drawn a diagram and determined the volume of an arbitrary shell?
what's my mistake there? i solved it again and the result is the same.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
what's my mistake there? i solved it again and the result is the same.
\(\displaystyle 4^2-\left(3+x^2 \right)^2=16-\left(9+6x^2+x^4 \right)=7-6x^2-x^4\)

You see, you left off the exponent of the squared term.
 

paulmdrdo

Active member
May 13, 2013
386
oh yes. the answer should be 48pi/5.

the last problem could you show me how to set up the integral. i have no idea how to to that.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
oh yes. the answer should be 48pi/5.

the last problem could you show me how to set up the integral. i have no idea how to to that.
Start by doing a sketch...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
oh yes. the answer should be 48pi/5.

the last problem could you show me how to set up the integral. i have no idea how to to that.
Yes, correct for the first one. Now for the second, consider the following diagram:

paulmdrdo2.jpg

The region to be revolved is shaded in yellow, for an arbitrary shell, the radius $r$ is the horizontal line in green and the height $h$ is the vertical line in red. The thickness of the shell is $dx$. And so the volume of this element is:

\(\displaystyle dV=2\pi rh\,dx\)

If $x$ is the $x$-coordinate of the vertical line, then what is the radius when revolved about $x=-2$, i.e., what is the length of $r$? What is the length of $h$?

edit: I misread the lower limit, it should be -1 instead of 1, but the method is the same. :D
 

paulmdrdo

Active member
May 13, 2013
386
Yes, correct for the first one. Now for the second, consider the following diagram:

View attachment 1446

The region to be revolved is shaded in yellow, for an arbitrary shell, the radius $r$ is the horizontal line in green and the height $h$ is the vertical line in red. The thickness of the shell is $dx$. And so the volume of this element is:

\(\displaystyle dV=2\pi rh\,dx\)

If $x$ is the $x$-coordinate of the vertical line, then what is the radius when revolved about $x=-2$, i.e., what is the length of $r$? What is the length of $h$?

edit: I misread the lower limit, it should be -1 instead of 1, but the method is the same. :D
the radius is r=x+2 am i correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
the radius is r=x+2 am i correct?
Yes, that is correct:

\(\displaystyle r=|x-(-2)|=|x+2|\)

For $x\in[-1,2]$, we have $x+2>0$ so:

\(\displaystyle r=x+2\)

What is $h$?
 

paulmdrdo

Active member
May 13, 2013
386
h is $e^{2x}$

then the integral should be $\displaystyle 2\pi\int_{-1}^2 (x+2)e^{2x}dx$

is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
h is $e^{2x}$

then the integral should be $\displaystyle 2\pi\int_{-1}^2 (x+2)e^{2x}dx$

is this correct?
Yes, that is correct. (Sun)