# Area of region

• Oct 20th 2011, 02:08 PM
icelated
Area of region
Basically, what i have is

$V = 2 \pi \int _{o}^ {4} (4-x) \, \sqrt x \, dx$

would i then distribute?

$V = 2 \pi \int _{o}^ {4} (4x^{1/2} -x ^{3/2} ) dx$

Then, integrate?

$2 \pi \int _{o}^ {4} (4 * \frac {2} {3} x^{3/2} - \frac {2}{5} x ^{5/2}$

then,

$2 \pi [ \frac {8} {3} x^{3/2} - \frac {2}{5} x ^{5/2}]$
evaluated from 0 to 4

Is this correct? I cant seem to get a decent answer. Wolfram gives me weird steps and an answer i cant match...And when would i do something with the 2pi? Do i use 2pi for each side?
• Oct 20th 2011, 02:33 PM
ebaines
Re: Area of region
Quote:

Originally Posted by icelated
$2 \pi [ \frac {8} {3} x^{3/2} - \frac {2}{5} x ^{5/2}]$
evaluated from 0 to 4

Is this correct?

Looks good so far. When you evaluate from 0 to 4 you should get:

$2 \pi [ (\frac 8 3 4 ^{3/2} - \frac 2 5 4^ {5/2} )-(0 - 0)] = \frac {256 \pi}{15}$
• Oct 20th 2011, 02:35 PM
TheEmptySet
Re: Area of region
Quote:

Originally Posted by icelated
Basically, what i have is

$V = 2 \pi \int _{o}^ {4} (4-x) \, \sqrt x \, dx$

would i then distribute?

$V = 2 \pi \int _{o}^ {4} (4x^{1/2} -x ^{3/2} ) dx$

Then, integrate?

$2 \pi \int _{o}^ {4} http://mathhelpforum.com/calculus/190884-area-region.html#post691275\$
$2 \pi [ \frac {8} {3} x^{3/2} - \frac {2}{5} x ^{5/2}]$

This means

$2 \pi [ \frac {8} {3} x^{3/2} - \frac {2}{5} x ^{5/2}] \bigg|_{0}^{4} =2\pi\left[ \frac{8}{3}4^{3/2}-\frac{2}{5}4^{5/2}-\left( \frac{8}{3}0^{3/2}-\frac{2}{5}0^{5/2}\right) \right]$

Now just use algebra to simplify
• Oct 20th 2011, 03:15 PM
icelated
Re: Area of region
That answers my questions thank you. Do i need to do anything to the thread?
• Oct 20th 2011, 05:35 PM
Ackbeet
Re: Area of region
Quote:

Originally Posted by icelated
Do i need to do anything to the thread?

Nope. We don't generally close threads around here unless there's been a violation of rules. Although, come to think of it, if you like, you can mark the thread as solved. That's in the "Thread Tools" menu.