Integration application: solids of revolution

[paulmdrdo]

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!

 

[Petrus]

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 \(\displaystyle \int_a^b\)

Spoiler

solve \(\displaystyle 3+x^2=4\)

 

[MarkFL]

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]

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).

 

[MarkFL]

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]

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]

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]

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]

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]

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:



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.

 

[paulmdrdo]

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.

the radius is r=x+2 am i correct?

 

[MarkFL]

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]

h is $e^{2x}$

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

is this correct?

 

[MarkFL]

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.