Help on intergration (1 Viewer)

[kittyful]

really hahaha, but i already have a tutor

 

[Drongoski]

really hahaha, but i already have a tutor

But is he as good as bof ? Is so why do you keep posting straightforward questions here??

 

[kittyful]

But is he as good as bof ? Is so why do you keep posting straightforward questions here??

because the questions look to striaghtforward for them to answer

 

[Drongoski]

because the questions look to striaghtforward for them to answer

NO. You pay them; it is their job to help you. That's what they are paid for. Don't feel bad asking them - otherwise what do your tutors do for you. Is this a 1-to-1 or a coaching centre. I'm sorry but I do not mean to run you down. Trying to be helpful if I can.

 

[kittyful]

a centre

 

[Drongoski]

How do i find the area enclosed by the curvees y= x^2 and x=y^2

Here you have 2 parabolae, one upright and the other sideway. If you sketch their graphs you can see they intersect at 2 points. You can find these by equating:






Edit

Kittyful: As a matter of curiosity, without naming the centre: how many lessons/term; how mant hours/lesson & what do they charge? Maybe cheaper 1-on-1.

Some centres are excellent & set a hot pace - they are good only for those who can keep up.

 

[kittyful]

Here you have 2 parabolae, one upright and the other sideway. If you sketch their graphs you can see they intersect at 2 points. You can find these by equating:






Edit

Kittyful: As a matter of curiosity, without naming the centre: how many lessons/term; how mant hours/lesson & what do they charge? Maybe cheaper 1-on-1.

Some centres are excellent & set a hot pace - they are good only for those who can keep up.

2 hours a week once a week idk how much they charge ... can't remember

 

[Drongoski]

2 hours a week once a week idk how much they charge ... can't remember

Thank you. If you think it is benefitting you then fine. If you are trailing the class then a good 1-on-1 tutor may be the way to go.

 

[bored of fail 1]

lol damm tutoring colleges, impossible to compete with them, "they cost tons more so they must be better ":|

plus everyone gets into a tutoring college as soon as yr11 starts and then they are stuck there for the whole hsc because none can be bothered to change places/get private tutor or they feel obligated to stay with the place even when they are not getting what they want ( like kittyful above ). @Kittyful: if you dont feel comfortable asking these questions then it should be quite obvious that you are not in the right place.

I have been told about an online tutoring site that is meant to be good: pays $18.50/hr , you work from home and its purely like an msn tutoring sorta thing ( kinda like what I am doing on here, but more private, and you actually get paid for it ! , finally someone might give me a chance to earn a living off what I like the most, helping people with 2/3U maths questions because as it stands now I will never get anyone for private tutoring ) , you can pick when you work and can do as many or as few hours as you like a week.

It says I must have atleast 3years tutoring experience, but pfft, screw that, ill just make it up and say that I have been tutoring people.

 

[Tophel]

Anyone know how to integrate sqr(9-x^2)? Is it possible with 2/3u?

 

[Bored Of Fail]

substitute

x= 3sin(a)

its 3unit, in 3unit they would give you the substitution

 

[kittyful]

Uploaded with ImageShack.us

Can anyone help with question 2 ... i tried it many times ... and some how my answers never match the answer at the back T_T ... ps. do i feel confortable in the tutor i am in now, its just that with the amount of students ... its quite difficult to ask questions ....

 

[bored of fail 1]

is your scanner drunk, I can just make out the words, only just though

and you obviously arent comfortable with the tutor if u have to ask all the questions on here

 

[kittyful]

lol i cbb to use a scanner ... i took a photo of it using the webcam .. haha i'll fix that



Uploaded with ImageShack.us

Q2bi the curve is y=x^1/3

 

[bored of fail 1]

2. a

i.

first note it has x intercepts -2 and +2

so volume = pi integral y^2 dx
= pi integral ( x^2-4)^2 dx
= pi integral ( x^4 -8x^2 +16) dx [ between limtis -2..2 ]

you can do the rest , calculator work

Continuing on ....

ii.

draw up the picture and you will see that the lower limit of the intergal is -4 ( the minium value of x^2 -4 )

and the upper limit of the integral is y=0 ( ie the x axis )

now volume about y axis is pi integral x^2 dy

so vol = pi integral ( y +4 ) dy [ between -4.. 0 ]

you can do calculations.

 

[MrBrightside]

lol i cbb to use a scanner ... i took a photo of it using the webcam .. haha i'll fix that



Uploaded with ImageShack.us

Q2bi the curve is y=x^1/3

lol I was gona say how could a scanner be so fucked up, but you said webcam so that explains it.

 

[bored of fail 1]

2 b .

i. Volume = pi integral y^2 dx = pi integral ( x^(1/3))^2 dx [ between x=0..x=8 ] ( know your index laws )
= pi integral ( x^(2/3) ) dx [ x=0..8 ]

ii.

Volume = pi integral x^2 dy

 

[bored of fail 1]

2b ii is a different type of question, and illustrates why you should always draw the graph

 

[kittyful]

2. a

i.

first note it has x intercepts -2 and +2

so volume = pi integral y^2 dx
= pi integral ( x^2-4)^2 dx
= pi integral ( x^4 -8x^2 +16) dx [ between limtis -2..2 ]

you can do the rest , calculator work

Continuing on ....

ii.

draw up the picture and you will see that the lower limit of the intergal is -4 ( the minium value of x^2 -4 )

and the upper limit of the integral is y=0 ( ie the x axis )

now volume about y axis is pi integral x^2 dy

so vol = pi integral ( y +4 ) dy [ between -4.. 0 ]

you can do calculations.

2a i got the exact same working out ... but somehow ... it didn't match lol

 

[bored of fail 1]

note the actual area you are rotating about the y axis, the volume is actually a cylinder of height 2 and radius 8 but from this cylinder you then subtract the volume of the curve about the y axis ( between y=0 and y=2 )

Volume of cylinder = pi r^2 h

so Volume = pi ( 8^2 ) (2 ) - pi integral (x^2) dy [ integrated from y=0..y=2 ]

now x^(1/3) =y
so x= y^3
x^2 = y^6

so volume = 128 pi - pi integral (y^6) dy [ from y=0..y=2]

you do calculations