Volumes by slicing (1 Viewer)

[vds700]

cambridge, ex 6.1 Q 3

The region [(x, y) : 0<_x<_2, 0 <_y<_2x - x^2] is rotated about the y-axis. Find the volume of the solid (slicing method)

cant understand the solution.

 

[undalay]

asdf

 

[conics2008]

sketch y=2x-x^2

take slices pd to y axis

let roots be x1 and x2

because you cant identify the length of the radius

since it form a doughnut shape your delta V becomes

pi(x^2 2 - x^2 1 ) dy

where x2=length of the big radius
x1=length of small radius

you will get a quadratic equation in the form of x^2-2x+y=0

use quad equation to find x2 and x1 which are your roots and remember the equation

hence your x2= 1+root of 1-y
x1= 1-root of 1-y

sub this into your delta v equation

lim deltha y >>0 sigma between 1 and 0 pi S 1 and 0 4root of 1-y

integrate and sub in limits.

edit you could of also used sum of roots and product of roots, but this method lets you play it safe because your find the exact root of the equation.

 

[vds700]

Thanks guys, makes sense now

 

[Js^-1]

The region bounded by the curve y=x(2-x), and the x-axis is rotated about the y-axis. Taking slices parallel to the x-axis, you find that the cross section of each slice is an annulus with radii x<SUB>1</SUB>, x<SUB>2</SUB> with x<SUB>2</SUB> > x<SUB>1</SUB>. Since x<SUB>1</SUB>, x<SUB>2</SUB> both lie on the curve y=x(2-x), they can be considered the roots of the equation as a quadratic in terms of x. [Note we normally find the roots of a quadratic when it is equal to zero, hence finding the zeros. ]
The annulus has volume δV=pi*(x<SUB>2</SUB>+x<SUB>1</SUB>)(x<SUB>2</SUB>-x<SUB>1</SUB>)δy.

To find x<SUB>2,</SUB>x<SUB>1</SUB> we need to solve the quadratic y=x(2-x).
x<SUP>2</SUP>-2x+y=0

x<SUB>2</SUB>+x<SUB>1</SUB>=2 [sum of roots = -b/a]
x<SUB>2</SUB>x<SUB>1</SUB>=y [product of roots = c/a]

Now (x<SUB>2</SUB>-x<SUB>1</SUB>)<SUP>2</SUP>=(x<SUB>2</SUB>+x<SUB>1</SUB>)-4(x<SUB>2</SUB>x<SUB>1</SUB>)
(x<SUB>2</SUB>-x<SUB>1</SUB>)<SUP>2</SUP>=2-4y
(x<SUB>2</SUB>-x<SUB>1</SUB>)=√(2-4y)

Hence the volume of the annulus is
δV=[pi*2√(2-4y) ] δy.

Note the δy indicates the integration is with respect to y.

Now
V= Lim<SUB>δy--> 0</SUB>S{y=0,1}[pi*2√(2-4y) ] δy.

V= 2pi ∫{0,1} √(2-4y)dy


V= 2(√2)*pi ∫ √(1-2y)dy
Let u = 2y
du = 2dy

at y=0 u=0
y=1 u=2

V= (√2)*pi ∫ √(1-u) du
V=(√2)*pi [-2/3(1-x)<SUP>3/2</SUP>]{0,2}
V=-2/3(√2)*pi [0-2]
V=(4√2*pi)/3 Units<SUP>3</SUP>
<SUP></SUP>
<SUP></SUP>
I probably made a few mistakes but i think thats a way to do it...

[Edit]: Forgot to change the limits of integration for the original, cant be bothered finding the other mistakes tonight.

 

[vds700]

The region bounded by the curve y=x(2-x), and the x-axis is rotated about the y-axis. Taking slices parallel to the x-axis, you find that the cross section of each slice is an annulus with radii x₁, x₂ with x₂ > x₁. Since x₁, x₂ both lie on the curve y=x(2-x), they can be considered the roots of the equation as a quadratic in terms of x. [Note we normally find the roots of a quadratic when it is equal to zero, hence finding the zeros. ]
The annulus has volume δV=pi*(x₂+x₁)(x₂-x₁)δy.

To find x_2,x₁ we need to solve the quadratic y=x(2-x).
x²-2x+y=0

x₂+x₁=2 [sum of roots = -b/a]
x₂x₁=y [product of roots = c/a]

Now (x₂-x₁)²=(x₂+x₁)-4(x₂x₁)
(x₂-x₁)²=2-4y
(x₂-x₁)=√(2-4y)

Hence the volume of the annulus is
δV=[pi*2√(2-4y) ] δy.

Note the δy indicates the integration is with respect to y.

Now
V= Lim_{δy--> 0}S{y=0,1}[pi*2√(2-4y) ] δy.

V= 2pi ∫{0,1} √(2-4y)dy


V= 2(√2)*pi ∫ √(1-2y)dy Let u = 2y
du = 2dy

V= (√2)*pi ∫ √(1-u) du
V=(√2)*pi [-2/3(1-x)^3/2]{0,1}
V=-2/3(√2)*pi [0-1]
V=(2√2*pi)/3 Units³


I probably made a few mistakes but i think thats a way to do it...

Thanks man, mustve taken ages to type that out. I think u made a mistake somewhere, the answer is 3pi/8, have a look at the way undalay did it, his method was slightly different, but got the right answer, i cant see whats wrong with your though.

 

[Js^-1]

Yeh it took me freaking forever. I might try and do the question tomorow in a free and see where i made a mistake.

[Edit]: I forgot to change the limits of integration...ill fix it now and see if it changes anything.

 

[Js^-1]

Ok I did it today and found out where i went wrong.

The region bounded by the curve y=x(2-x), and the x-axis is rotated about the y-axis. Taking slices parallel to the x-axis, you find that the cross section of each slice is an annulus with radii x<SUB>1</SUB>, x<SUB>2</SUB> with x<SUB>2</SUB> > x<SUB>1</SUB>. Since x<SUB>1</SUB>, x<SUB>2</SUB> both lie on the curve y=x(2-x), they can be considered the roots of the equation as a quadratic in terms of x. [Note we normally find the roots of a quadratic when it is equal to zero, hence finding the zeros. ]
The annulus has volume δV=pi*(x<SUB>2</SUB>+x<SUB>1</SUB>)(x<SUB>2</SUB>-x<SUB>1</SUB>)δy.

To find x<SUB>2,</SUB>x<SUB>1</SUB> we need to solve the quadratic y=x(2-x).
x<SUP>2</SUP>-2x+y=0

x<SUB>2</SUB>+x<SUB>1</SUB>=2 [sum of roots = -b/a]
x<SUB>2</SUB>x<SUB>1</SUB>=y [product of roots = c/a]

Now (x<SUB>2</SUB>-x<SUB>1</SUB>)<SUP>2</SUP>=(x<SUB>2</SUB>+x<SUB>1</SUB>)<SUP>2</SUP>-4(x<SUB>2</SUB>x<SUB>1</SUB>) [I forgot to put the squared in (x<SUB>2</SUB>+x<SUB>1</SUB>)<SUP>2</SUP>]
(x<SUB>2</SUB>-x<SUB>1</SUB>)<SUP>2</SUP>=4-4y
(x<SUB>2</SUB>-x<SUB>1</SUB>)=√(4-4y)

Hence the volume of the annulus is
δV=[pi*2√(4-4y) ] δy.

Note the δy indicates the integration is with respect to y.

Now
V= Lim<SUB>δy--> 0</SUB>S{y=0,1}[pi*2√(4-4y) ] δy.

V= 2pi ∫{0,1} √(4-4y)dy


V= 2(2)*pi ∫ √(1-y)dy



V=4*pi [-2/3(1-y)<SUP>3/2</SUP>]{0,1}
V=-8/3*pi [0-1]
V=8*pi/3 Units<SUP>3</SUP>

hmmm, the answer was 3*pi/8 you say...well beats me. I'll ask my teacher where i went wrong.

 

[tommykins]

The answer is 8pi/3.

 

[vds700]



hmmm, the answer was 3*pi/8 you say...well beats me. I'll ask my teacher where i went wrong.

no you are right, when i typed my last post i tried to remember the answer, but got it wrong- my bad.

Thanks everyone for their help.

 

[Js^-1]

No worries, glad I could help. Good practice too.