thewisestudent9
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99 POINT QUESTION PLUS BRAINLIEST!!!
(Please answer genuinely, and do not answer just for point, if you do your answer will be deleted, and the points you earned, will be taken away...)
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Using the Disk, Shell, or Washer method from Calculus...

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the y-axis...

4.) 49-x^2; y=0
A.) V = 4pi * integral of(0 to 7) [ x(49-x^2) dx = 2401pi
B.) V = 2pi * integral of(0 to 7) [ x(49-x^2) dx = (2401/2)pi
C.) V = 2pi * integral of(-7 to 7) [ x(49-x^2) dx = (2401/2)pi
D.) V = 4pi * integral of(-7 to 7) [ x(49-x^2) dx = 2401pi

**Please explain your answer as much as possible (if you could, please involve graphs), if you do not your explain how you got your answer, it will be deleted, and the points you earned will be taken back...
**REMINDER: This is Calculus, not your basic Math...

Respuesta :

apologiabiology
yay, I like das stuff


I think I learned the washer method
goes like this


[tex]area=\pi \int\limits^a_b {(outercurve)^2-(innercurve)^2} \, dx [/tex]

ok, so

actuallly, this is easier

y=49-x^2 and
y=0
see when they intersect again
they intersect at -7 and 7
if we do integrate from -7 to 7, then it wil give 0 (because integration is area under the curve), so note that they are same both sides so integrate from 0 to 7 then double the volume to get both sides
so it can't be C or D
It also can'nt be A because it should not be multipied by 4, it should be multipied by 2


basically, we don't need the washer method
remember, area=pir^2
the disk method then
we are summing up all the radii disks and squareing them
but doubleing them so
[tex]area=2\pi (\int\limits^7_0 {49-x^2} \, dx )^2 [/tex]
no idea why we need the inside part to be x(49-x^2), that is intersting

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