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Mathematics 17 Online
OpenStudy (kayders1997):

Please help volume between two curves: y=1/4x^2 and y=5-x^2

OpenStudy (kayders1997):

@zepdrix @jim_thompson5910 @satellite73

OpenStudy (kayders1997):

@FortyTheRapper

OpenStudy (kayders1997):

@vishweshshrimali5

OpenStudy (fortytherapper):

I rather use my computer than mobile phone. Let me check if it's being used

OpenStudy (kayders1997):

Okay

OpenStudy (kayders1997):

First I drew a sketch

OpenStudy (fortytherapper):

It is. Let me boot it up and I'll be here

OpenStudy (kayders1997):

I drew a sketch than gave up 😂

OpenStudy (anonymous):

is it \[y=\frac{1}{4}x^2\] and \[y=5-x^2\]?

OpenStudy (kayders1997):

Yeah

OpenStudy (anonymous):

do you know where they meet up? you can kinda do it in your head

OpenStudy (kayders1997):

2

OpenStudy (anonymous):

if not, this nice picture will help http://www.wolframalpha.com/input/?i=5-x^2,x^2%2F4

OpenStudy (kayders1997):

So x=4

OpenStudy (anonymous):

no at \(x=-2\) and at \(x=2\)

OpenStudy (anonymous):

if that is not clear, solve \[\frac{1}{4}x^2=5-x^2\] for \(x\) put both \(-2\) and \(2\) get you \(1\)

OpenStudy (kayders1997):

Oh my bad was looking at wrong problem

OpenStudy (kayders1997):

But yes that makes sense

OpenStudy (anonymous):

since evidently \(5-x^2\) is larger, integrate \[\int_{-2}^2(5-x^2-\frac{1}{4}x^2)dx\]

OpenStudy (kayders1997):

Okay

OpenStudy (fortytherapper):

The legend explained it perfectly

OpenStudy (kayders1997):

Lol sorry internet problems

OpenStudy (anonymous):

lol oh just one more thing this is area, not volume

OpenStudy (kayders1997):

Oh

OpenStudy (fortytherapper):

How's the integration?

OpenStudy (kayders1997):

Well

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