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a level maths core 3 question solids of revolution

Question: The region n R bounded by the curve y=x((x^3)+1)^3, the x-axis between 0 and 1, and the line x=1, is rotated through 2pi radians about the x-axis. Find the volume of the solid of revolution formed.
How would you go about the question, you`ll have to get the equation into y^2 first, but how would you do that? expand all the brackets or am i missing something?
Reply 1
Original post by itlstallion69
Question: The region n R bounded by the curve y=x((x^3)+1)^3, the x-axis between 0 and 1, and the line x=1, is rotated through 2pi radians about the x-axis. Find the volume of the solid of revolution formed.
How would you go about the question, you`ll have to get the equation into y^2 first, but how would you do that? expand all the brackets or am i missing something?


try the substitution of u=x^3+1
Original post by itlstallion69
Question: The region n R bounded by the curve y=x((x^3)+1)^3, the x-axis between 0 and 1, and the line x=1, is rotated through 2pi radians about the x-axis. Find the volume of the solid of revolution formed.
How would you go about the question, you`ll have to get the equation into y^2 first, but how would you do that? expand all the brackets or am i missing something?


No need to expand.

y2=x2(x3+1)6y^2=x^2(x^3+1)^6 and hence the substitution u=x3+1u=x^3+1 does the job when integrating it.
Someone link me to the fancy handwriting symbols page thingy
Original post by Bulletzone
Someone link me to the fancy handwriting symbols page thingy


lol what is that?
Reply 5
Original post by Bulletzone
Someone link me to the fancy handwriting symbols page thingy

Fancy handwriting symbols page thingy
am coming to an answer of 128pi all over 21 but the book says 127pi all over 21
Reply 7
Original post by itlstallion69
am coming to an answer of 128pi all over 21 but the book says 127pi all over 21


you may have forgotten to minus the 1pi/27 from the second part of the indefinite integral
Original post by yusyus
you may have forgotten to minus the 1pi/27 from the second part of the indefinite integral


yeh just noticed i disregarded the last part of the integral since the x value was zero, thanks

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