Volumes of solids of revoultion question (1 Viewer)

[kangarulz]

*help* Integration

Need help with working out:

Find the area bounded by the curved y=(9-x)^1/2 and the x and y axes.
Find the volume of the solid generated when the area is rotated about:
(i)the x axis
(ii) the y axis


Show that y=2(1-x^2)^1/2 and y=(1-x^2)^1/2 intersect at (1,0) and (-1,0). Find the volume of the resultant solid when the enclosed region is rotated about:
(i)the x axis
(ii)the y axis


Find the volume of the solid when the region between the curves y=x₂ and y=2-x₂ in the x-y plane is rotated about:
(i) the x axis (ii) the y axis

If the curve xy=1 is rotated about the x axis, find the volume generated by the arc of the curve intercepted between the planes x=1 and x=4

 

[kangarulz]

bump

 

[pLuvia]

Re: *help* Integration

(i)
pi∫{-3-->3}(9-x)²dx
=pi∫{-3-->3}(81-18x+x²)dx
=pi[81x-9x²+x³/3]{-3-->3}
=pi|-504|
=504pi u³

(ii)
pi∫{0-->9}(9+y²)²dy
...

Can you work for there? Be back later

 

[kangarulz]

Re: *help* Integration

(i)
pi∫{-3-->3}(9-x)²dx
=pi∫{-3-->3}(81-18x+x²)dx

Where u have (9-x)² is actually:

{(9-x)^1/2}² hence, the square root of (9-x)

 

[Riviet]

Re: *help* Integration

Need help with working out:

Find the area bounded by the curved y=(9-x)^1/2 and the x and y axes.
Find the volume of the solid generated when the area is rotated about:
(i)the x axis
(ii) the y axis

Area bounded by curve and x and y axes=
9
∫ (9-x)^1/2dx=[-2/3.(9-x)^3/2]⁹₀
0

= 0 - [-2/3.27]

= 18 units²

(i) Volume=
9
∫pi.[(9-x)^1/2]²dx=
0

9
∫pi.(9-x)dx=pi[9x - x²/2]⁹₀
0
=pi[(81 - 81/2)-0]
=(81pi)/2 units³

(ii) Rearrange the curve to get it in terms of x:

x=9-y², limits are y=0 and y=3.

.'. Volume=
3
∫pi.(9-y²)²dy=
0

3
∫pi.(81-18y²+y⁴)dy=pi.[81y-6y³+y⁵/5]³₀
0

=pi.[(243 - 162 + 243/5)-0]

=(648pi)/5 units³

P.S Please clarify what you mean by "y-2(1-x^2)^1/2" as being a function.

 

[kangarulz]

sorry that shouldve been y=2(1-x²)^1/2

 

[pLuvia]

Re: *help* Integration

Opps, mistake

 

[kangarulz]

Re: *help* Integration

hey can i just ask, with:

9
∫ (9-x)1/2dx=[-2/3.(9-x)3/2]90
0

= 0 - [-2/3.27]

= 18 units2


why is it a -2/3?

 

[pLuvia]

Re: *help* Integration

Because he integrated the inside as well, as part of the reverse chain rule in integration

 

[Riviet]

Re: *help* Integration

Show that y=2(1-x^2)^1/2 and y=(1-x^2)^1/2 intersect at (1,0) and (-1,0). Find the volume of the resultant solid when the enclosed region is rotated about:
(i)the x axis
(ii)the y axis

Solving the two equations simultaneously:

2.sqrt(1-x²)=sqrt(1-x²)
4(1-x²=1-x²
4-4x²=1-x²
3x²=3
x=+1
Substituting into either equation gives y=0.

.'. cuves intersect at (1,0) and (-1,0)

i) Volume=
1
pi ∫[2.sqrt(1-x²)]² - [sqrt(1-x²)²dx=
-1

1
pi∫3(1-x²)dx
-1

The rest should be fairly straight forward.

ii) First, we express each equation in terms of x:
y=sqrt(1-x²) becomes x=sqrt(1-y² and y=2.sqrt(1-x²) becomes x=sqrt(1 - y²/4)

At this point, you might want to sketch each curve to determine which volume to subtract. The unit circle is simple enough, but the other is an ellipse which you do not need to know how to sketch in the 2/3 unit course. It's basically a oval shaped curve with x-intercepts 1 and -1 and y-intercept 2. From your sketch, the ellipse is outside circle, so we subtract the circle's volume from the ellipse's to find the enclosed volume between the two curves. We also need to note the difference in limits for each curve, for the circle it's simply y=0 and y=1 but for the ellipse, it's from y=0 to y=2.

.'. Volume=
2
pi∫[sqrt(1 - y²/4)]²dy - pi[0->1]∫[sqrt(1-y²)]²dy

You can finish it off.

 

[Riviet]

Re: *help* Integration

If the curve xy=1 is rotated about the x axis, find the volume generated by the arc of the curve intercepted between the planes x=1 and x=4

xy=1
y=1/x
y=x^-1, with limits x=1 and x=4.

.'. Volume=

4
pi∫(x^-1)²dx
1

You can do the rest.

 

[Riviet]

Re: *help* Integration

Find the volume of the solid when the region between the curves y=x₂ and y=2-x₂ in the x-y plane is rotated about:
(i) the x axis (ii) the y axis

I think you meant the curves y=x² and y=2-x² as there's no enclosed region between two linear functions. You must have confused the sup tag with sub.

First we find intersection point of two curves:
x²=2-x²
x²=1
x=+1
So our limits are x=1 and x=-1.
Again, drawing a sketch helps you picture it and work out which curve has a greater area between it and the x-axis. In this case it's the y=2-x² curve.

i) Volume=
1
pi∫[(2-x)² - (x²)²]dx
-1

ii) For this, we need to split the curve in the line y=1, since the areas are different.
You need to draw a diagram for this.
Rearranging the two equations in terms of x,
y=x² becomes x=sqrt(y) and y=2-x² becomes x=sqrt(2-y).
To find the volume of the enclosed area rotated about the y-axis, we can add the volume formed from rotating x=sqrt(y) about y-axis with limits y=0 and y=1 to the volume formed from rotating x=sqrt(2-y) about y-axis with limits y=1 to y=2.

.'. Volume=
1
pi∫[sqrt(y)]²dy + pi[1->2]∫[sqrt(2-y)]²dy
0