integration (1 Viewer)

[fullonoob]

y = x^3 + 2 y = 3x
Region bounded by the curves and the y axis.
calculate the volume of the solid generated when this region is rotated about y axis
Intersects at (1,3) and (-2,-6)
blonde moment maybe

 

[MetroMattums]

Er... can you rephrase this question?

There are two equalities in a line... ><

 

[fullonoob]

Er... can you rephrase this question?

There are two equalities in a line... ><

there was individual parts so i tried to simplify it. I dont know where to take the integral of a and b at. This is the whole question
1) Find the equation of the tangent to the curve y = x^3 +2 at the point where x= 1
2) draw a diagram showing the region bounded by the curve, the tangent and the y axis.
3) calculate the volume of the solid generated when this region is rotated about the y axis

it was too tedious to type up so i simplified it. Obviously have to find point of intersection

 

[yugi]

there was individual parts so i tried to simplify it. I dont know where to take the integral of a and b at. This is the whole question
1) Find the equation of the tangent to the curve y = x^3 +2 at the point where x= 1
2) draw a diagram showing the region bounded by the curve, the tangent and the y axis.
3) calculate the volume of the solid generated when this region is rotated about the y axis

it was too tedious to type up so i simplified it. Obviously have to find point of intersection

Damnit, i tried to use the stupid latex thing and failed twice and wasted alot of time haha sorry you are just going to have to read this

1) y=x^3+2
y'=3x^2
when x=1
y'=3

when x=1, y=(1)^3+2=5
.'. the point is (1,5)
using the formula: y-y1=m(x-x1)
we get: y-5=3(x-1)
y-5=3x-3
.'. 3x-y+2=0 is the equation of the tangent

2)You will need to draw the curve and tangent then solve the two simultaneously to find the points of intersection which will become your boundaries a and b

3)Use the formula
V=<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 10"><meta name="Originator" content="Microsoft Word 10"><link rel="File-List" href="file:///C:%5CDOCUME%7E1%5Cyuxi%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman";} </style> <![endif]-->∏∫x^2dy

 

[fullonoob]

Damnit, i tried to use the stupid latex thing and failed twice and wasted alot of time haha sorry you are just going to have to read this

1) y=x^3+2
y'=3x^2
when x=1
y'=3

when x=1, y=(1)^3+2=5
.'. the point is (1,5)
using the formula: y-y1=m(x-x1)
we get: y-5=3(x-1)
y-5=3x-3
.'. 3x-y+2=0 is the equation of the tangent

2)You will need to draw the curve and tangent then solve the two simultaneously to find the points of intersection which will become your boundaries a and b

3)Use the formula
V=<META content=Word.Document name=ProgId><META content="Microsoft Word 10" name=Generator><META content="Microsoft Word 10" name=Originator><LINK href="file:///C:%5CDOCUME%7E1%5Cyuxi%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml" rel=File-List><STYLE> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </STYLE>∏∫x^2dy

incorrect. when x=1, y=(1)^3+2=5 already wrong.
answer is 2/5 pi u^3. Lol i know the steps, its just that i dont get the answer.

 

[fullonoob]

bump for more replies xD
Gna go do work soon...and still cant get the right answer

x = y/3 x^2 = y^2/9
x^3 = y-2 x^2 = (y-2)^(2/3)
V = pi integral 3 and 0 y^2/9 dy - pi integral 3 and 0 (y-2)^(2/3)
= pi [y^3/27 - 3/5(y-2)^(5/3)] between 3 and 0
doesn't get me the answer i need

 

[yugi]

oops LOL

 

[fullonoob]

bump again...not answered

 

[Trebla]

I'm going assume the area referred to is the one in x > 0.

bump for more replies xD
Gna go do work soon...and still cant get the right answer

x = y/3 x^2 = y^2/9
x^3 = y-2 x^2 = (y-2)^(2/3)
V = pi integral 3 and 0 y^2/9 dy - pi integral 3 and 0 (y-2)^(2/3)
= pi [y^3/27 - 3/5(y-2)^(5/3)] between 3 and 0
doesn't get me the answer i need

The lower limit should be 2. If you draw the diagram, the curve passes the y-axis at y = 2 and thus there is no existing "volume" below y = 2 for that cubic cube, otherwise you are considering the area on the other side of the y-axis.

 

[fullonoob]

I'm going assume the area referred to is the one in x > 0.


The lower limit should be 2. If you draw the diagram, the curve passes the y-axis at y = 2 and thus there is no existing "volume" below y = 2 for that cubic cube, otherwise you are considering the area on the other side of the y-axis.

oh damn got mixed up with the point of intersection
i'll check it out now

 

[Trebla]

I probably should have clarified that the lower limit in the second integral only (i.e. of the cubic curve) is 2. The first integral still has a lower limit of 0.

If you think about it graphically, the volume is found by the volume of the cone of radius 1 and height 3 (when integrating to find the volume enclosed by the linear function) and then subtracting off the volume found by rotating the area bounded by that cubic curve, the y-axis and the line y = 3 about the y-axis.

 

[fullonoob]

I probably should have clarified that the lower limit in the second integral only (i.e. of the cubic curve) is 2. The first integral still has a lower limit of 0.

If you think about it graphically, the volume is found by the volume of the cone of radius 1 and height 3 (when integrating to find the volume enclosed by the linear function) and then subtracting off the volume found by rotating the area bounded by that cubic curve, the y-axis and the line y = 3 about the y-axis.

i see the cone now :L
but dont know the integrals still
cubic has upper lower 0 2??
linear has upper lower 0 3 ??
clarify again please

 

[Trebla]

Think about what these integrals represent diagrammatically on the graph...