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Baka may sagot kayo badly needed pls

Lmr123

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Find the volume of the solid revolution generated by the curves y=x2 and the line x=3 and the x-axis y revolving around the line x=-3

asap
 
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
 
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
sure na to lods? sasagot ko na to HAHAHAHHA
 
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