Volumes Q (1 Viewer)

[Makematics]

so guys, this is an 8 mark volumes question 8 from a past trial from my school. my school often misallocates marks, but this was a shocker...
anyways for some reason my answer is slightly different from the solution, and it is driving me nuts because i cant find any mistakes in my work or the solutions.
Find the volume of the solid generated by rotating the region common to the circles x^2 +y^2 = 16 and x^2 + y^2=8x about their common chord.

i know, it is insane that 8 marks would be given for such a simple question...

 

[Sy123]

The areas on either side of the chord are the same, also the area underneath the x-axis is same as above x-axis, so we can split the volume into a quarter piece which is the volume generated by rotating the region from x=2 to x=4 and y=0 and y=sqrt(16-x^2)

using shells



Is this correct?

 

[Makematics]

The areas on either side of the chord are the same, also the area underneath the x-axis is same as above x-axis, so we can split the volume into a quarter piece which is the volume generated by rotating the region from x=2 to x=4 and y=0 and y=sqrt(16-x^2)

using shells



Is this correct?

no i think delta V should be 4pi times the stuff in the brackets, though i guess it depends on what your delta v is. i did shells too, but the solution did it by slices so its hard to compare.
yet your volume is double the volume in the solutions. i think you took a quarter instead of one half. since you are rotating t about x=2, the rotation already covers it if that makes sense.
i think im integrating it wrong... but i cant see where

 

[Sy123]

no i think delta V should be 4pi times the stuff in the brackets, though i guess it depends on what your delta v is. i did shells too, but the solution did it by slices so its hard to compare.
yet your volume is double the volume in the solutions. i think you took a quarter instead of one half. since you are rotating t about x=2, the rotation already covers it if that makes sense.
i think im integrating it wrong... but i cant see where

Yep that was silly of me

For integrating its best to expand, the first term xsqrt(16-x^2) is easily dealt with reverse chain rule/easy sub, the second one is done better geometrically by taking the appropriate segment of the circle.

 

[Makematics]

Yep that was silly of me

For integrating its best to expand, the first term xsqrt(16-x^2) is easily dealt with reverse chain rule/easy sub, the second one is done better geometrically by taking the appropriate segment of the circle.

yep i split it, i think im stuffing up the first one. what do you get for 4pi times integral of x sqrt(16-x^2)dx

 

[Makematics]

OMG GOOD GOD... WORST MISTAKE... on the very last line... trust me to check it over like 10 lines and not pick it up. alright well thanks for helping