Math problem

[Mashed-Potatos]

I'm having some trouble with my calculus homework. Hopefully someone can help me out.

sqrt() = square root

Consider the solid obtained by rotating the region bounded by the given curves about the line y = 5.

y=5x, y=5*sqrt(x)

Find the volume V of this solid.

Here's what I started to do:

I used washers to find the volume.
x=y/5, x=sqrt(y/5)

Pi*[integral from 0 to 5] (5-y/5)^2 - (5- sqrt(y/5))^2 dy
= Pi*[integral from 0 to 5] (25-2y/5+(y^2)/25)-(25+10*sqrt(y/5)-y/5) dy
= Pi*[integral from 0 to 5] (y^2)/25 - 3y/5 +10*sqrt(y/5) dy
= Pi*((y^3)/75 - (3y^2)/10 +(4/3)*sqrt(5)*y^(3/2)
and I ended up with (55*Pi)/2

[shnedit]

you are not worthy for me to share my great knowledge of maths.

[Saturday Saint]

That looks right.

[Mashed-Potatos]

Well I had to have fucked up somewhere because when I press the submit button it's telling me it's wrong.

[Attau]

integral from 0 to 5 :sly:?

[Mashed-Potatos]

integral from 0 to 5 :sly:?

I was thinking that might be the problem, but since I was using y as the variable I thought that would be right. Should it be from 0 to 1?

[Attau]

And if you rotate about y=5 why did you write your equations in terms of y?

Your cross section should be parallel to the y-axis so you should integrate with respect to x....

[Ozzy Wrong]

Your hypermeter rhetoric is off. Stabalize it, then we'll talk.

[Lacap11]

...math

[Attau]

Your hypermeter rhetoric is off. Stabalize it, then we'll talk.

Hahaha. I have no idea what you just said but I enjoyed it.

[Saturday Saint]

Your hypermeter rhetoric is off. Stabalize it, then we'll talk.

Shut up and go mess with your damn gigabytes.

@Attau: Ozzy didn't say anything that was supposed to make sense.

[Attau]

Here's what you do....

you rotate about y=5 and draw a cross section.

The outer rim of the cross section is y=5x and your inner rim is y=5sqrt(x).

The cross section is a circle so.... Pi(5-5x)^2 is the area of a cross section at any given point (for the outer rim) minus Pi(5-5sqrt(x))^2 at any given point to remove the area between 5 and the inner ring of the solid object.

so the equation is Pi* [integral] (5-5x)^2 - (5-5sqrt(x))^2 from 0 to 1 (where the equations 5x and 5sqrt(x) are equal).

/thread

[Attau]

Shut up and go mess with your damn gigabytes.

@Attau: Ozzy didn't say anything that was supposed to make sense.

Yea I guessed that :ninja:. Just enjoyed it.

[Mashed-Potatos]

Since I'm spinning it around a horizontal line, shouldn't I write the equation with y? At least that's what I have written in my notes...

[Ozzy Wrong]

Hahaha. I have no idea what you just said but I enjoyed it.
You're not at the level of technological enlightenment that I have reached.
Shut up and go mess with your damn gigabytes.

@Attau: Ozzy didn't say anything that was supposed to make sense.

Unfortunately, the neutron processor I obtained from my school tech room doesn't collaborate with my current gigabytes. I think the problem is that I left one of my Playboy magazines in the internet last time I was there.
Tweety.

[Attau]

Well.... I can't remember all the cases that are possible.... But read the post and think about the logic. *shrug*

[stalwart]

forumfall is NOT your personal army.

[shnedit]

forumfall is NOT your personal army.

im dissapointed at all the serious replies:(

[Attau]

I'm sorry forumfall :(. I just like math

[Morthor]

I'm sorry forumfall :(. I just like math

whats WRONG with you???!

[Triffe]

Since I'm spinning it around a horizontal line, shouldn't I write the equation with y? At least that's what I have written in my notes...

Depends on the method you are using, washer or shell. If your using shell, it is the same as the axis your rotating about, so if you rotate about the y-axis, you use y, washer would have you use x in this case.

[Mashed-Potatos]

Alright, just tried it how you said Attau. I ended up with (50*Pi)/3 and it said it was wrong. Then I tried one more time (without factoring out a 5) and got 55*Pi/6, too bad I used up all my attempts on that question.
Thanks for the help though.

Depends on the method you are using, washer or shell. If your using shell, it is the same as the axis your rotating about, so if you rotate about the y-axis, you use y, washer would have you use x in this case.
Yeah I just got them mixed up.

[Ozzy Wrong]

Alright, just tried it how you said Attau. I ended up with (50*Pi)/3 and it said it was wrong. Then I tried one more time (without factoring out a 5) and got 55*Pi/6, too bad I used up all my attempts on that question.
Thanks for the help though.


Yeah I just got them mixed up.

The hyphen in your name makes you seem exotic and intelligent.
I'm attracted.

[Attau]

Alright, just tried it how you said Attau. I ended up with (50*Pi)/3 and it said it was wrong. Then I tried one more time (without factoring out a 5) and got 55*Pi/6, too bad I used up all my attempts on that question.
Thanks for the help though.


Yeah I just got them mixed up.

Pi*(25/6)

You prolly had a problem factoring through -1 in the extended integral?

[iza]

Ah this problem reminds me of how much I hated Calc II last year.
I wish I could say I would never have to look at this shit ever again, but unfortunately I have to take differential equations next semester... </rant>

Note: I could fairly easily solve the OP problem, but I won't, because fuck you.

[Attau]

Ah this problem reminds me of how much I hated Calc II last year.
I wish I could say I would never have to look at this shit ever again, but unfortunately I have to take differential equations next semester... </rant>

Note: I could fairly easily solve the OP problem, but I won't, because fuck you.

Oh yea I forgot to add that.


Fuck you OP!

[Mashed-Potatos]

I still have to take Calc 2 next semester. But hey, fuck you too.

[Viare]

I'm having some trouble with my calculus homework. Hopefully someone can help me out.

sqrt() = square root

Consider the solid obtained by rotating the region bounded by the given curves about the line y = 5.

y=5x, y=5*sqrt(x)

Find the volume V of this solid.

Here's what I started to do:



fuck you, you bitch. Are you tryin to make my brain blow up. Im gona pk you.

[Mashed-Potatos]

The hyphen in your name makes you seem exotic and intelligent.
I'm attracted.

Your Nov. 2008 join date turns me on. We should get together sometime.

[alfaroverall]

Since I'm spinning it around a horizontal line, shouldn't I write the equation with y? At least that's what I have written in my notes...
If you wanted to you could integrate with respect to either variable, it will just change the way you actually do the integrand (and in many cases one route will lead you to a non-algebraic antiderivative). But doing it the other way (with a cross section being parallel to your axis of rotation) uses the shell method instead of the disk/washer method.

Anyhow, I think people have pretty much led you in the right direction so I won't bother to try to explain stuff.