Let's see if Sup Forums is smart enough to answer this easy calculus question:
What is the area under the curve of x^2 rotated around the y axis from bounds x = 0 to x = 5?
Let's see if Sup Forums is smart enough to answer this easy calculus question:
What is the area under the curve of x^2 rotated around the y axis from bounds x = 0 to x = 5?
This is a volume Q not area you faggot. Go back to your cave from whence you came
Wouldn't that be a volume?
Yo know what I mean you niggers
Let's see if OP is smart enough to ask an easy calculus question.
the same as if it was rotated from 0 to -5
That would be a cup with a very thin bottom and edges sharp as fuck.
OP here, will give you guys a hint since you guys are fucking retarded. Use the shell method.
Brainlet, if you're in calc 2, "you know what I mean" doesn't fucking fly. You need to learn to be way more detail obsessive if you you want to have any hope of surviving your fucking sophomore year.
lrn2/sci/
Is that Calc 2? I'm in Calc 1 right now I didn't know integrals applied to volumes too
shell method is for calc 2 faggots.
>calling the people he's asking to do his homework niggers.
>Basic Calculus
Google it or shove it
Easy F(b)-F(a)
pi * integral[0 to 25] of (5 - rooty)^2 dy
that's the equation i think i don't want to do any more
Integral, at its very basic, is an infinite sum of infinitesimal pieces. With area, we use rectangles, with volumes we use various different methods, one of which is rectangles rotated about an axis to make hollow cylinder.
You guys are purposely being obtuse to avoid answering the question. Be condescending all you want, but this is a basic integral that all engineering students know and will take about 2 minutes to solve.
Wrong
seeing as im retarded ill go the rout of actually answering
if we assume y=x^2
The bounds of y are 0 to 25
So...
Integral(1 dy from 0 to 25)*integral(x^2 dx) from 0 to 5
25*integral(x^2 dx) from 0 to 5
25*1/3(5^3) = 25/3*125
Calculated out to 1041.6
I thought it was 25pi/2 units^3
Can you end up finding the volume of the hollow cylinder through a geometric formula as well?
Incorrect
Please retake Calc 2
41.7
>still insulting the people he's asking for help with his basic math homework.
It's Christmas Eve and my term ended 2 weeks ago. You really think this is homework?
I learned this shit in calc 1
I think you're trying to find out why you failed your final.
let's do the multivariablefag way, just for shits and giggles
area under f(x, y) = x^2 + y^2 on the circle of radius 5
in polar, f(r, theta) = r^2
so, theta goes from 0 to 2pi, and r goes from 0 to 5
so, (integral of dtheta from 0 to 2pi)*(integral of r^2 from 0 to 5) = 2pi*((5^3)/3) = 250pi/3
3272.5
your answer is 1/2 the correct answer - shells take 2pi, not just pi - just pi is for washers
he was mostly right, give him a break
Am I right or wrong F(b)-F(a) right? Not too sure what you mean by rotated around the y axis
...wrong
when the graph is rotated 360 degrees around the y-axis, it forms a solid with rotational symmetry around the origin
in OP's example, imagine you have a bowl on a table - we're supposed to find the volume between the bowl and the table
there are special formulas you have to use when graphs are rotated around axes; they're called the washer and shell methods, and you pick one for a problem
Yes, it'd be the bigger volume minus the hollow volume.
Volume of a cylinder is area x height, area being (pi)r^2, thus h(pi)r^2. Since the only difference between the cylinders is radius you'd get
h(pi)r1^2 - h(pi)r2^2 = pi(h(r1^2-r2^2))
With Cal 2 you actually end up with many cylinders of varying heights which is where integrals and whether you use dx or dy comes in.
OP here
This is the correct answer
testing myself
> early 30s, haven't used calculus since college a decade ago
find integral of x^2, which is 1/3x^3
plug in 5, find value
plug in 0, find value
get difference
answer
or did i do something wrong?
nope, that's not the way to do it
you have to use the washer method or the shell method, if you're sticking to calc 2
Is it 625pi?
OP asked for the volume when x^2 is revolved around the y axis
pi* integral (from 0 to 5) [(x^3)/3]^2
pi* ((5^3)/3)^2
Final Answer: 5454
I learnt this just last semester. 1st year eng at McMaster University
I'm mildly retarded
the polar coord transformation puts an extra r, so dx dy becomes r dr dtheta
with this in mind, the integral is r^3, not r^2, from 0 to 5
then I get OP's answer
ooohhhhh
missed that part
there's a multiply by pi somewhere in there then
If you wanted to make it hard, you lazy fuck, why did you make the y-axis a bound? And why did you use such a baby-tier function?
Anyone who remembers anything about rotational volumes is going to be able to solve this in a heartbeat. At least swim in the parametric end of the kiddie pool.
winner
>can't even ask the question correctly
>wants Sup Forums to solve a calc I problem
"HUUUUUURRRRRR, answer this question HURRRRRRR, ghghlshdlfhlsh Sup Forums!!!!"
Jesus christ, some next level retard.
oh come on, I nearly had it in but fucked up the transformation and corrected myself in
Wait so it's like a 3D model? Isn't that multi variable calc?
>Wow I'm so smart doing what a calculator does but less efficiently
stop waisting your brainpower
you're a faggot, it's wrong
That shit's wrong
OP proved that Sup Forums couldn't solve an introductory Calc II question
>mfw
lol you are seriously trying to gloat because you learned some basic calc and a random handful of Sup Forumstards either don't know the answer or can't remember it off-hand?
wtf is wrong with you kid?
A calculator cannot do calculus lol
It turns out you don't need multivariable calc to do this problem because of the rotational symmetry. They invented the brainless "washer" and "shell" methods to let people who never passed calc 3 solve these types of problems and feel smart. Of course if you know multivariable the problem is trivial in cylindrical coordinates.
How do you pass Calc II but couldn't pass Calc III???
why do shells take 2pi? I thought that it was just pi
...
>call it CALCULator
>it can't do CALCULus
Seriously? What cheap piece of shit are you using??
treating it like a 3D model is one inefficient way to do it
the way you're supposed to do it with only calc 2 is with OP's method, shown in funnily enough, you can derive the general formula OP used from my solution. For whatever reason, I'll write it out.
You have a function f(x) that you're revolving around the y-axis and integrating from x=a to x=b. Plugging in r, the radius (because we're in polar mode now), we get f(r).
In polar coords, we have:
integral from 0 to 2pi and from a to b of f(r) dr dtheta
This turns into:
Integral from a to b of 2pi * f(r) * r * dr
This becomes 2pi times the integral of x*f(x) from a to b, which is OP's formula.
I think using the shell method is easier than using multivar
y axis not x axis
Pi*integral 0 to 5 [(x^2)]^2
(x^4)
X^(4+1)/4+1
X^5/5
V=pi*[(5^5/5)-0] =625pi
wrong axis you mong
OP said the y-axis, not the x-axis
you're correct for the x-axis tho
Ha ha you're wrong ha
Try being 19 somewhere else
shells take 2pi, washers take pi.
see my "proof" in
Dude, it's about the Y AXIS... lol
that's 2 times the correct answer
I have no idea how you got your original [(x^2)]^2 expression, so can't help you
Only because you either just learned multivar or haven't learned it at all. After a few years it's just annoying to remember all these specialized methods; actual mathematicians would never use the shell method, they would just construct the triple integral because that is a more general and powerful way of solving these types of problems.
Wait a minute op said area not volume
you're right, if only because multivar includes a lot of machinery that you won't need for this problem
sorry! fixed it
he corrected himself in , responding to , who had the same issue
OP is retarded.
See
What machinery? It's actually just knowing the cylindrical Jacobian determinant, there's nothing complicated about using multivar for this.
No, "actual mathematicians" would use the most direct and easiest way of solving the problem.
You yourself even messed up your multivar solution, albeit you did correct yourself shortly thereafter
Just use 625pi, math professors like that better for some reason.
you're using the wrong axis for the washer/disk method
with washers, you're rotating around the same axis that you're integrating with.
In the problem, the rotation happens around the y-axis, so a washer solution should involve dy, not dx
if it were dx, the revolution would be around the x-axis instead
How did you start with x^2^2?
Op's answer is correct
I was the dude with the (initially) fucked-up multivar solution, not him
just thought you should know
sorry, machinery was the wrong word. I was talking more about multiple integrals - in calc 2, you only know how to do single integrals and shit yourself every time there's more than one integral sign on the right.
Not sure who you think I am but I didn't post any attempt at a solution in this thread. And a triple integral is the most direct and easiest way; the shell and washer methods are shitty constructs that are only useful before you understand multiple integration.
OP thinks he's smart but I bet he can't calculate the integral from 0 to infinity of e^(-(x)^2)dx
What caused you to post an incorrect solution in the first place? What were you trying to accomplish?
triple? I'm pretty sure a double integral in polar is fine, since we have rotational symmetry. (See my answer in and correction in )
Fair enough, but I think the shell method and multivar are equally as complicated to use for this problem.
just an oversight on my part, absolutely unintentional
idiot kys calc2 literally easiest shit in the world
YOU CANT JUST SOLVE A TRIPLE INTEGRAL ARE YOU CRAZY NIGGER
sure, agreed.
No, in this situation, integrating x^3 and plugging in the bounds is loads easier than setting up a triple integral.
The shell method was made for questions like this, why not utilize it?
>SOLVE
evaluate? how's it hard?
It is a triple integral since your answer has dimensions of (length)^3. The rotational symmetry just makes the angular integral trivial, as you pointed out.
fucking y axis. lmfao
>The shell method was made for questions like this
because pointless arguments are fun, I'll point out that the shell method was made so that students who can only do single integrals can solve questions like this
The general consensus is that calc 2 is the hardest of calc's, including elementary differential equations.
Christ, you attempted like 4 times, I would never want to hire you as an engineer
i have the squared because what you do is. You use the x in the function as the radius. and we know that pi*r**2 is the area of a circle. So basically what we do is, we make a shit ton of circles and add the areas of all of em, and that's the volume. I learnt this last semester. 1st year Engineering
...
nigger what
if you're using washers, the outside is the cylinder of radius 5 and the inside is the paraboloid, so in the integral sign you'd have 5^2 - (sqrt(y))^2
i know lmfao, this is final though
the only correct solution