Question

Find the volume of the solid obtained by rotating the region under the curve y= (3/x+3) over the interval [0, 1] about the y-axis.

Answer 1

\[2\pi\int\limits_{0}^{1}x\frac{3}{x+3}dx\]

Answer 2

is there any website you can show that simplifies that a wittle further for me (:

Answer 3

zarkon provided you what is sometimes called "the shell method"

Answer 4

the shell method can make life easier ... in this case we could integrate by parts in one or two steps to get the results

Answer 5

\begin{array}c
--&--&\frac{3}{x+3}\\
+&x&3\ ln(x+3)\\
-&1&\int3\ ln(x+3)
\end{array}

which may or maynot be easier

perhaps a usub technique?

Answer 6

\[\frac{3x\ dx}{x+3}\]
u = x+3
du = dx

x = u-3
\[\int \frac{3(u-3)\ du}{u}\]

Answer 7

this amounts to:

\[3\left(\int\frac{u}{u}du-3\int\frac{1}{u}du\right)\]
\[3\left(u-3\ ln(u)\right)\]
\[3u-9\ ln(u)\]
\[3(x+3)-9\ ln(x+3)\]
\[3x+9-9\ ln(x+3) \text{ ; from 0 to 1}\]

Answer 8

9+9-9 ln(4)) - ( 9 - 9 ln(3))

18 -9 ln(4) - 9 + 9 ln(3)

9 -9 ln(4) + 9 ln(3) perhaps?

Answer 9

missed it someplace; i get 6 more than wolfram

Answer 10

i see it maybe; I did 3(3) instead of 3(1) :)

3(1) +9 -9ln(4) - (3(0) +9 -9ln(3)) ... thats better

Answer 11

ohhhhh. Whats the final answer? lol

Answer 12

lol ... its:

http://www.wolframalpha.com/input/?i=int%283x%2F%28x%2B3%29%29dx+from+0+to+1

Answer 13

hmm.. thats weird.. My choices are merely decimals of 2. something

Answer 14

p.s. you sent me the solution to y= 3x/x+3 when it was x/x+3 >:| Thanks for that.

Answer 15

\[x\frac{3}{x+3}=3\frac{x}{x+3}=3\frac{x+3-3}{x+3}=3\frac{x+3}{x+3}+3\frac{-3}{x+3}=3+\frac{-9}{x+3}\]

then integrate

Answer 16

Huh?