volumes integration help plz! (1 Viewer)

[koolkid59]

Find the volume of the soild generated by rotating the region bounded by the following curves about .....

the x-axis:
c) y = 5x-x^2 and y = 0
d) y = x^3-x and y = 0

the y-axis:
a) x = y - 2, y = 1, and x = 0

plz show working clearly....
thanks in advance!

 

[bleakarcher]

You have to be specific with d) in terms of which quadrant the region is located in.

 

[SpiralFlex]

c)

Factorising,

Of course this shape comes to no surprise. It is a parabola.



So we want to rotate the region enclosed between the graph and the line (In other words, the axis.). It is also good to visualise the shape when rotated especially in later HSC Mathematics.

The shape is a CD PlAYER FROM SANTA.











d)



It is a cubic!

Of course straight away you should realise that this cubic is actually an odd function. (Symmetrical about the origin) We can use such a powerful knowledge to simplify our integrals. I think it's good to show this however.


Let







Hence the function is an odd function.


Draw it!



If we said it has symmetry about the origin, then both parts are equal in volume. So we need to find one volume and just double it! You will find the shape to be TWO DISTORTED TENNIS BALLS! (Close to it anyway.)











e) Drawing another diagram, of course this shape is linear!













Alternatively, this shape is an CORNETTOS ICE-CREAM CONE!

 

[bleakarcher]

Spiralflex, the question was on volume, not area.

 

[SpiralFlex]

Spiralflex, the question was on volume, not area.

BLERRRRRRGGGGGHHHHHHHH!

 

[bleakarcher]

Find the volume of the soild generated by rotating the region bounded by the following curves about .....

the x-axis:
c) y = 5x-x^2 and y = 0
d) y = x^3-x and y = 0

the y-axis:
a) x = y - 2, y = 1, and x = 0

plz show working clearly....
thanks in advance!

c) V=pi*integral[y^2] dx from x=0 to x=5
V=pi*integral[25x^2-10x^3+x^4] dx from x=0 to x=5
V=pi*[(25/3)x^3-(5/2)x^4+(1/5)x^5] from x=0 to x=5
d) V=2pi*integral[y^2] dx from x=0 to x=1 since y is odd
V=2pi*integral[x^6-2x^4+x^2] dx from x=0 to x=1
V=2pi[(1/7)x^7-(2/5)x^5+(1/3)x^3] from x=0 to x=1
a) V=pi*integral[x^2] dy from y=1 to y=2
V=pi*integral[y^2-4y+4] from y=1 to y=2
V=pi[(1/3)y^3-2y^2+4y] from y=1 to y=2

Evaluate.

 

[Carrotsticks]

BLERRRRRRGGGGGHHHHHHHH!

lol'd hard.

 

[Lemiixem]

Can someone help my understand how:
x^3 = y turns into x = y^1/3
Similar to that how would I make x by itself in this equation:
x^3 = y - 2 turns into x = !?!??!!

 

[SpiralFlex]

Can someone help my understand how:
x^3 = y turns into x = y^1/3
Similar to that how would I make x by itself in this equation:
x^3 = y - 2 turns into x = !?!??!!

I think you mean make the subject of the formulae.



The reason is because if we cube something, to get the variable by itself, we need to cube root it. So that is why we cube root both the LHS and RHS.


The same argument goes for the next question.





Rearranging this to index form,

 

[wilsondw]

Spiralflex, the question was on volume, not area.

I dun get it..how was his working out not related to volume?

He used pi didnt he? or she

 

[SpiralFlex]

I dun get it..how was his working out not related to volume?

He used pi didnt he? or she

I accidentally mistook the question for area. Did not read it properly. I have edited the post.