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sammie

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I was wondering if anyoen could please explain ot me how to do this question. And give me pointers on how to do these in the future, ta.

Question:
The region enclosed between the curve y=x^3, the x axis and the lines x=1 x=2 is rotated about the x-axis. Calculate the volume of the solid so formed.
Thank you:p
 

Winston

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When something is rotated about the x-axis then the equation's subject must be

y^2

and if it's about the y-axis it the subject is

x^2


so in this case let's get our equation first


y=x^3
y^2 = (x^3)^2
= x^6

and we know our x values which is 1 and 2

so let's form the entire question out


please note ~ is the integral sign


Pi 1~2 (x^6) dx

------------------ 2
= Pi [ x^7 / 7 ]
------------------1

= Pi ((2)^7/7) - ((1)^7/7)

= Pi (128/7) - (1/7)

= Pi (127/7)

thus = 127Pi/7 Cubic Units
 

Supra

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yeh u gotta remeba that when its rund the x axis its y and round the y axis, its x (with volumes)
 

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