Question

Let R be the region between the graph of f and the x-axis on the interval . Find the volume V of the solid of revolution generated by revolving R about the x-axis. Give both exact (symbolic) and approximate (numeric) answers
f(x)=8x-x^2

Answer 1

so to find this i squared the function, multiplied it by pi, and found the indefinite integral of this from 0 to 8

Answer 2

the from 0 to 8 did not end up here when i copied and pasted it sorry

Answer 3

so i got pi(8x^2-x^4)
then i got 8pix^2-pix^4

Answer 4

the indefinite integral i got from that is
(8/3)pix^3 - (pi/5)x^5

Answer 5

i then plugged in 8 and plugged in 0 and subtracted 0 from 8 so i got
[(8pi/3)8^3 - (pi/5)8^5] - [(8pi/3)0^3 - (pi/5)0^5]
which gets simplified to
(8pi/3)8^3 - (pi/5)8^5
which gets simplified to
(4096pi/3) -(32768pi/5)

Answer 6

but its wrong

Answer 7

(8x-x^2)^2 is not 8x^2-x^4

Answer 8

its not?

Answer 9

(a+b)^2 = a^2 +2ab +b^2

Answer 10

oh i have to foil?

Answer 11

so it is 64x^4-16x^6+x^8?

Answer 12

\[\large (8x-x^2)^2=64x^2-16x^3+x^4\]

Answer 13

you dont add the exponents? when multiplying

Answer 14

so the antiderivative is
(64*pi/3)x^3 - (16*pi/4)x^4 +(pi/5)x^5 correct?

Answer 15

yes

Answer 16

100% correct

Answer 17

so when i put that equation in just plug in 8 for all the x why is it wrong?

Answer 18

(64*pi/3)8**3-(16*pi/4)8**4+(pi/5)8**5 i put that in to be exact

Answer 19

the ** represents exponents

Answer 20

since when you plug in 0 to x it equals 0 i did not subtract it, just plugged in 8 for x..correct?

Answer 21

yes

Answer 22

so do you know why it marked me wrong?

Answer 23

i have 1 more chance

Answer 24

no.

Answer 25

probably calculation error

Answer 26

i didnt calculate it i plugged it in like this
(64*pi/3)8**3-(16*pi/4)8**4+(pi/5)8**5

Answer 27

but its ok i used wolfram so i dont know how to do it but i got credit for it on my hw thx guys!