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Y=X, Y=X^(.5) rotate about x=2
Find the volume...

given answering is 8pi/15
however I keep getting -7pi/15

Thanks for much for help !

2007-09-02 17:26:48 · 4 answers · asked by edbiology 1 in Science & Mathematics Mathematics

4 answers

i used shell method

2pi [integral 0-1] [(2-x)(x^(1/2) -x)] dx =
2pi [integral 0-1] [2x^(1/2) - 2x - x^(3/2) + x^2] dx =
2pi [(4/3)x^(3/2) - x^2 - (2/5)x^(5/2) + (1/3)x^3] (from 0-1)
2pi [(4/3) -1-(2/5)+(1/3)] =
2pi (4/15) = 8pi/15

2007-09-02 17:39:32 · answer #1 · answered by eman 2 · 0 0

Unclear what region is being rotated about the line x=2. One thing is sure: the volume cannot be negative.

2007-09-03 00:37:05 · answer #2 · answered by ironduke8159 7 · 0 0

Y=X whch the .5 and X=Y^.5, you would divide the Y=X with the .5 and subtract the one from x, making it 1x and then add the .5 which makes it 2.25.....duh!

2007-09-03 00:31:51 · answer #3 · answered by Anonymous · 0 1

totally right.

the "book [or given] answers" can sometimes be wrong
or maybe you made a mistake you aren't aware of

good luck with that

it's just a question, probably on a homework assignment or something?

You'll get over it.

2007-09-03 00:31:57 · answer #4 · answered by Dovie ❊ 6 · 0 2

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