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.
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
hmm i get 3pi but 3pi isn't in the choices
@Jhannybean
my answer choices are a. 15/2 pi b. 16pi c. 17/2pi d. 11pi
@Jhannybean
@Luigi0210
Not volume D:
and the reason being cause Miss @Jhannybean made a booboo
ohh ok
sooo can u help?
Do you see where she took the anti-derivative and got x^2+4x?
yes
Well she forgot to put it over 2 :P because \[\int\limits xdx=\frac{ x^2 }{ 2 } +C\]
Now solve it and see what you get
Did you get it? @onegirl
yes sorry
so x^2 + 4x^2 ? or
No you are using \[\Pi[\frac{ x^2 }{ 2 }+4x] \]
ohhh ok thanks :D
Yup, it was all thanks to @Jhannybean .. without her equation we'd be lost :l
yea BUT i get -27/2 and its not in my answer choices :/
How did you get that? I got something different
you did? :/ i used WA
WA?
wolfram alpha
\[\Pi([0]-[\frac{ (-3)^2 }{ 2 }+4(-3)]\]) \[\Pi [0-(\frac{ 9 }{ 2 }-12)]\] \[\Pi[0-(\frac{ -15 }{ 2 })]\] \[\frac{ 15 }{ 2 }\Pi\] And oh, maybe you should try using your brain.. those things are wrong at times
okay
Hope that made sense ^_^'
oh hahaha -_- i forgot the two... darn.
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