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?

Question

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?

amistre64 · Answer

you can shell it

amistre64 · Answer

2pi {S} x[sqrt(x)] dx ; [0,6]

anonymous · Answer

yeah, that's how i did it

anonymous · Answer

but the answer is a decimal: 35.27...

amistre64 · Answer

x[sqrt(x)] can be re written as:

sqrt(x^3) = x^(3/2) and integrate

amistre64 · Answer

it gonna be a decimal regardless; pi is irrational/transcendental

anonymous · Answer

I am using a math homework software that won't take decimals for this answer. Should I just do it by hand?

anonymous · Answer

like have the fractional, un-evaluated part as and answer?

amistre64 · Answer

if it dont take decimals, then use the exact version of it...

anonymous · Answer

I will try right now

amistre64 · Answer

2x^(5/2)
-------- * 2pi  at x = 6
      5

anonymous · Answer

that's the answer, but I need to evaluate at 6. and the computer won't take my decimal answer

amistre64 · Answer

4.pi. sqrt(6.6.6.6.6) = 4.pi.36 sqrt(6)
----------------       ------------
            5                              5

amistre64 · Answer

144pi sqrt(6)
-----------  should be an exact answer then
         5

anonymous · Answer

you can evalutate it that way???

amistre64 · Answer

yeah; thats what it equate to right?

anonymous · Answer

yes, and it takes that answer! Thanks everyone!

amistre64 · Answer

:)  just glad I was right lol  :)

watchmath · Answer

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 · Answer

okay, so I can see how you can do that a little more mathematically

anonymous · Answer

:)

anonymous · Answer

Thanks!