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Mathematics 8 Online
OpenStudy (anonymous):

A region bounded by f(x) = sqrt(x + 4) , y = 0, x = -3, and x = 0 is shown below. Find the volume of the solid formed by revolving the region about the x-axis.

OpenStudy (jhannybean):

since it's bound by x=-3 and x=0, y=0 \[\large f(x)=\sqrt{x+4}\]\[\large y=\sqrt{x+4}\]\[\large A=\pi(\sqrt{x+4})^2\]\[\large A=\pi(x+4)\]Now we're finding volume,and you can find it by integrating this area we've found. \[\large V=\pi \int\limits_{-3}^{0}Adx\]\[\large V=\pi \int\limits_{-3}^{0}(x+4)dx\]\[\large V=\pi[x^2+4x]\] integrate from -3 to 0\[\large V=\pi[((0)^2+4(0))-((-3)^2+4(-3))]\]Simplify

OpenStudy (anonymous):

hmm i get 3pi but 3pi isn't in the choices

OpenStudy (anonymous):

@Jhannybean

OpenStudy (anonymous):

my answer choices are a. 15/2 pi b. 16pi c. 17/2pi d. 11pi

OpenStudy (anonymous):

@Jhannybean

OpenStudy (anonymous):

@Luigi0210

OpenStudy (luigi0210):

Not volume D:

OpenStudy (luigi0210):

and the reason being cause Miss @Jhannybean made a booboo

OpenStudy (anonymous):

ohh ok

OpenStudy (anonymous):

sooo can u help?

OpenStudy (luigi0210):

Do you see where she took the anti-derivative and got x^2+4x?

OpenStudy (anonymous):

yes

OpenStudy (luigi0210):

Well she forgot to put it over 2 :P because \[\int\limits xdx=\frac{ x^2 }{ 2 } +C\]

OpenStudy (luigi0210):

Now solve it and see what you get

OpenStudy (luigi0210):

Did you get it? @onegirl

OpenStudy (anonymous):

yes sorry

OpenStudy (anonymous):

so x^2 + 4x^2 ? or

OpenStudy (luigi0210):

No you are using \[\Pi[\frac{ x^2 }{ 2 }+4x] \]

OpenStudy (anonymous):

ohhh ok thanks :D

OpenStudy (luigi0210):

Yup, it was all thanks to @Jhannybean .. without her equation we'd be lost :l

OpenStudy (anonymous):

yea BUT i get -27/2 and its not in my answer choices :/

OpenStudy (luigi0210):

How did you get that? I got something different

OpenStudy (anonymous):

you did? :/ i used WA

OpenStudy (luigi0210):

WA?

OpenStudy (anonymous):

wolfram alpha

OpenStudy (luigi0210):

\[\Pi([0]-[\frac{ (-3)^2 }{ 2 }+4(-3)]\]) \[\Pi [0-(\frac{ 9 }{ 2 }-12)]\] \[\Pi[0-(\frac{ -15 }{ 2 })]\] \[\frac{ 15 }{ 2 }\Pi\] And oh, maybe you should try using your brain.. those things are wrong at times

OpenStudy (anonymous):

okay

OpenStudy (luigi0210):

Hope that made sense ^_^'

OpenStudy (jhannybean):

oh hahaha -_- i forgot the two... darn.

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