anonymous
  • anonymous
if you are given the terms y = sqrt(x), y = 0, and x = 6, how do you find the volume of the solid generated by revolving it about the y-axis?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
you can shell it
amistre64
  • amistre64
2pi {S} x[sqrt(x)] dx ; [0,6]
anonymous
  • anonymous
yeah, that's how i did it

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anonymous
  • anonymous
but the answer is a decimal: 35.27...
amistre64
  • amistre64
x[sqrt(x)] can be re written as: sqrt(x^3) = x^(3/2) and integrate
amistre64
  • amistre64
it gonna be a decimal regardless; pi is irrational/transcendental
anonymous
  • anonymous
I am using a math homework software that won't take decimals for this answer. Should I just do it by hand?
anonymous
  • anonymous
like have the fractional, un-evaluated part as and answer?
amistre64
  • amistre64
if it dont take decimals, then use the exact version of it...
anonymous
  • anonymous
I will try right now
amistre64
  • amistre64
2x^(5/2) -------- * 2pi at x = 6 5
anonymous
  • anonymous
that's the answer, but I need to evaluate at 6. and the computer won't take my decimal answer
amistre64
  • amistre64
4.pi. sqrt(6.6.6.6.6) = 4.pi.36 sqrt(6) ---------------- ------------ 5 5
amistre64
  • amistre64
144pi sqrt(6) ----------- should be an exact answer then 5
anonymous
  • anonymous
you can evalutate it that way???
amistre64
  • amistre64
yeah; thats what it equate to right?
anonymous
  • anonymous
yes, and it takes that answer! Thanks everyone!
amistre64
  • amistre64
:) just glad I was right lol :)
watchmath
  • watchmath
Let me just use the washer method for comparison \(\int_0^{\sqrt{6}}\pi(36-y^4)\,dy=\pi(36y-\frac{y^5}{5})\mid_0^\sqrt{6}=\pi(36\sqrt{6}-\frac{36\sqrt{6}}{5})=\frac{144\pi\sqrt{6}}{5} \)
anonymous
  • anonymous
okay, so I can see how you can do that a little more mathematically
anonymous
  • anonymous
:)
anonymous
  • anonymous
Thanks!

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