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🎓 Academic Baka may sagot kayo badly needed pls
- Thread starter Lmr123
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- badly needed needed sagot
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PHC- Naksu
Elite
Try lang lods
Intersection point of the two curves are by solving the two equations together as (-2,5) and (1,2)
Volume generated about x axis is∫πy^2 dx for each curve and by difference for region in between
=∫π(x^2+1)^2 dx- ∫π(-x+3)^2 dx range of x= -2 to 1
=π∫(x^4+2x^2+1)dx-π∫(x^2-6x+9)dx
=π(x^5/5+2x^3/3+x-(x^3/3-6x^2/2+9x) range of x= -2 to 1
=π(x^5/5+x^3/3+3x^2-8x) range of x= -2 to 1
=π((-2)^5/5+(-2)^3/3+3(-2)^2-8(-2))- (((1)^5/5+(1)^3/3+3(1)^2-8(1))
=π((-32/5-8/3+12+16)-((1/5+1/3+3-8))
=π(-33/5+30)
=π(117/5) =23.4π
Intersection point of the two curves are by solving the two equations together as (-2,5) and (1,2)
Volume generated about x axis is∫πy^2 dx for each curve and by difference for region in between
=∫π(x^2+1)^2 dx- ∫π(-x+3)^2 dx range of x= -2 to 1
=π∫(x^4+2x^2+1)dx-π∫(x^2-6x+9)dx
=π(x^5/5+2x^3/3+x-(x^3/3-6x^2/2+9x) range of x= -2 to 1
=π(x^5/5+x^3/3+3x^2-8x) range of x= -2 to 1
=π((-2)^5/5+(-2)^3/3+3(-2)^2-8(-2))- (((1)^5/5+(1)^3/3+3(1)^2-8(1))
=π((-32/5-8/3+12+16)-((1/5+1/3+3-8))
=π(-33/5+30)
=π(117/5) =23.4π
sure na to lods? sasagot ko na to HAHAHAHHA
WiggLeZ_101
Forum Veteran
sure nayan sagot muna dali
PHC- Naksu
Elite
sabi ko nga try lang eh hehehe if d mo naintindihan ung sinagot ko wag mo n isulat dhil d mo rin nman maipapaliwanag
Q
Queen Buns
Wala ka ng no choice kaya isulat mo na 
Wala ka ng no choice kaya isulat mo na
sasagot ko na, salamat sa inyo guys balitaan ko kayo kung ano score ko HAHAHAHHA
PHC- Naksu
Elite
cge goodluck lods kaya mong ipasa yan n nkatakip ang isang mata hehe
ayan ang sound :>
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