2U math problems D: (volumes of a revolution+locus) (1 Viewer)

[Fozby]

The area in the first quadrant bound by y = (9 - x^2), y=0 and y=9 is rotated about the y axis. Calculate the volume of the solid generated.

I keep getting:

Pi * Integral between 9/0 (9-x^2)^2

= " 81 - 18x^2 + x^4 - 0

= " 81x - [18x^3]/3 + [x^5]/5 - 0

= Pi * 81(9) - [18(9)^3]/3 + [(9)^5]/5 - 0
Pi * (729) - (4274) + ([59,049]/5) - 0
= some number over 20,000

Textbook says =127.235

Also;

The point P(x,y) moves so that its distance PA from the point A(1,5) is always twice its distance PB from the point B(4,-1). Show that the equation of the locus is x^2 + y^2 -10x + 6y + 14 = 0. I understand this to be 2PB = PA.

(x-1)^2 + (y+5)^2 = 2* [ (x-4)^2 + (y+1)^2 ]

x^2 - 2x + 1 + y^2 - 10y + 25 = 2* [ x^2 - 8x + 16 + y^2 + 2y + 1 ]

" = 2x^2 - 16x + 32 + 2y^2 + 4y + 2

x^2 - 2x + 1 + y^2 - 10y + 25 - 2x^2 + 16x -32 - 2y^2 - 4y - 2 = 0

-x^2 - y^2 + 14x - 14y - 4 = 0


Any and all help with either questions is appreciated.

 

[undalay]

the bounds for the first question isn't 0-9
Its 0-3

edit for q2

2PB = PA
4PB^2 = PA^2
thats where u went wrong.

 

[Fozby]

I don't understand how the boundary changes from 0-9 to 0-3. Can you explain that?

I also subbed in 3 for 9

= Pi * [ 81(3) - [18(3)^3]/3 + [(3)^5]/5 - 0 ]
= Pi * [ (243) - (162) + ([243]/5) - 0 ]
= 407.150

 

[undalay]

sorry i read the question wrong.

the bounds are 0->9
However sinces its rotated about the y axis
Your integrating x^2, not y^2
so u rearrange hte equation to
x^2 = 9 - y

 

[Fozby]

Ahh jesus i'm an idiot. I've done a thousand questions around the Y axis and I just got stumped :[ Thanks for the help

 

[tommykins]

The area in the first quadrant bound by y = (9 - x^2), y=0 and y=9 is rotated about the y axis. Calculate the volume of the solid generated.

Since its about the y axis, you make x the subject and square that then integrate it.

y = 9-x²
x² = 9-y

V = pi int 9 - y dy 9->0
V = pi[9y - y²/2] 9->0
Subbing 9 in (0 yields a 0 result) you get pi[81-81/2] = 40.5pi = 127.23 (2dp)

The point P(x,y) moves so that its distance PA from the point A(1,5) is always twice its distance PB from the point B(4,-1). Show that the equation of the locus is x^2 + y^2 -10x + 6y + 14 = 0. I understand this to be 2PB = PA.

Firstly, draw a diagram so you can visualise it.

sqrt [ (x-1)²+(y-5)² ] = 2sqrt [ (x-4)² + (y+1)² ] => squaring both sides.
[ (x-1)²+(y-5)² ] = 4[ (x-4)² + (y+1)² ] => expanding
x² - 2x + 1 + y² - 10y + 25 = 4[x²-8x+16 + y²+2y+1]
x² - 2x + 1 + y² - 10y + 25 = 4x² - 32x + 64 + 4y² + 8y + 4 => all to one side.
3x² - 30x + 3y² + 18y + 42 = 0 => divide everything by 3

x^2 + y^2 -10x + 6y + 14 = 0