Volume of Solids: Integration Question (1 Viewer)

[>> Lisa <<]

Hey, I am not sure how to approach this question.

Find the volume of revolution generated by revolving:
the curve y= (sq. rt) x from x=0 to x=4 about the line y=5.

Any help would be much appreciated. thanks.

 

[zenger69]

well to find the volume of anything you square the function

there y=x
ignore the y=5 because it's the same as around the x-axis (i think)
logically finding a solid rotated about y=5 is the same as a solid rotated about the x axis (y=0)

chuck a pi in front

so becomes V=pi (integral between 0 and 4) x dx

and find the volume.

 

[DistantCube]

Are you sure, Zenger? if you draw the function and rotate it around the x axis and then around y=5, you'll find they look different....

rotating the function y=sqrt x about y=5 is the same as rotating the function y=(sqrt x)-5 about the x axis.

So our new function is y=(sqrt x)-5 and you want to rotate it about the x axis.

therefore; V = pi times the integral from 0 to 4 of ((sqrt x)-5)^2 .dx

expand the brackets, integrate it (reverse chainrule wont work, its not a linear function), you get; [1/2x^2-10/(3/2) x^(3/2) +25x] between 0 and 4.

And I think it comes out to be something like 54 and two thirds.

 

[zenger69]

ok then... i might be wrong. I haven't come across any of those questions.

Remember I only do 3U

 

[velox]

Remember I only do 3U

Dont EVER say that again. Have confidence in yourself. And "only 3u" pfft, makes no difference.

 

[Trev]

Yeah velox is right! 'only 3unit', this IS extension 1 work!

 

[DistantCube]

word to that. It wasn't anything major anyway, just draw the graph and you can see it.

 

[|Axis_]

Hey, I am not sure how to approach this question.

Find the volume of revolution generated by revolving:
the curve y= (sq. rt) x from x=0 to x=4 about the line y=5.

Any help would be much appreciated. thanks.

One way is to shift the y=sqrt(x) curve down by 5, so that you are revolving y=sqrt(x)-5 about the x-axis (which is y=0).

The answer is then: pi * integrate[(sqrt(x)-5)^2]dx from 0 to 4.

 

[DistantCube]

*annoyed*

which is what I wrote with working....unless you were clarifying...

 

[|Axis_]

oops! i'm really sorry DistantCube. i was a bit drunk last night. sorry sorry sorry.

 

[mouldy_bread]

word to that. It wasn't anything major anyway, just draw the graph and you can see it.

that was going to be my advice. Draw it. Thats about as far as I've gotten in the volume aspect.