Question

Just checking an answer...

Find the volume of the solid obtained by rotating the region bounded by \(y = \sqrt{25 - x^2}, \quad y = 0, \quad x = 2, \quad and \quad x = 4\) about the x - axis.

is the answer \(74\pi\)??

Answer 1

\(\int_{2}^{4}\sqrt{25-x^2}dx\)?

Answer 2

yep that's what i used...this problem is not washer right?

Answer 3

i got something else... @lgbasallote

Answer 4

aww

Answer 5

@dpaInc , did I do the reduction right?

Answer 6

@inkyvoyd there was also a square

Answer 7

Yes, there was. :S

Answer 8

isn't the integrand supposed to be f(x)^2 ?

Answer 9

\(\int_{2}^{4}-x^2+25 dx\)

Answer 10

still, no...
@inkyvoyd , that's it..

Answer 11

don't forget the pi though...

Answer 12

what did i do wrong =_=

Answer 13

apple preferably....

Answer 14

So, \(\frac{-x^3}{3}+25x|_{x=2}^{x=4}\)

Answer 15

I think you forgotten the pi too

Answer 16

No, I didn't. I made it invisible. -.-

Answer 17

like himself... :)

Answer 18

\[\pi | 25x - \frac{x^3}{3} |_2^4\]
\[\pi {[(100 - 50) - (\frac{-64}{3} - \frac{8}{3})]}\]

Answer 19

\[\pi [50 - (\frac{-72}{3})]\]

\[\pi[50 + 24] = 74\pi\]

Answer 20

I SEE WHAT I DID WRONG!!

Answer 21

@inkyvoyd , why did you take out your soln?

Answer 22

\[\frac{206\pi}{3}\] @dpaInc

Answer 23

Cause it was wrong.

Answer 24

then mine is wrong too... 94pi/3 ???

Answer 25

I got like
32 and 1/3 pi

Answer 26

but that's wrong i think

Answer 27

mine is 31 and 1/3 pi

Answer 28

oh yeah lol 94pi/3 hahaha

Answer 29

@dpaInc is right "S

Answer 30

change to spherical coordinates?

Answer 31

you found your err?

Answer 32

\[(100 - 50) - [\frac{64}{3} - \frac{8}{3}] = 50 - \frac{56}{3} = \frac{94}{3}\]

Answer 33

oh, the 64 was -64 before....

Answer 34

i double-negatived it haha

Answer 35

i don't never do that... :)

Answer 36

lol =))

well i gotta get home now..thanks :D

Answer 37

(btw..im in the university and it's like 7 pm here getting creepy)

Answer 38

roha...