[SOLVED] calculus volume (disk method)

**Legendsn3verdie** · September 12th 2008, 11:27 AM

can somebody let me know wehre i went wrong i doubt i got the right answer.. those dashes are really equal signs, i should of made it bigger.

**Legendsn3verdie** · September 12th 2008, 12:58 PM

you guys see where i have a y^7/2..

i got that by [ y^3/2)^2 ...

dont i add 3/2 + 4/2 to get that 7/2?? dont u add a power to a power.. or am i just supposed to mulitply?

**Soroban** · September 12th 2008, 01:11 PM

Hello, Legendsn3verdie!

Sorry, I don't understand any of your integrals . . .

Find the volume of: . $\begin{Bmatrix}x \:=\:y^{\frac{3}{2}} \\ x \:=\:0 \\ y\:=\:2 \end{Bmatrix}$ . revolved about the y-axis

Code:

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        |             *
       2+ - - - * - - -
        |:::*
        |:*
        |*
        |
    - - * - - - - - - - - -
        |

Formula: . $V \;=\; \pi\int^b_a \!\!x^2\,dy$

We have: . $V \;\;=\;\;\pi\int^2_0\!\! \left(y^{\frac{3}{2}}\right)^2\,dy \;\;=\;\; \pi\int^2_0\!\! y^3\,dy \;\;=\;\;\frac{\pi}{4}y^4\,\bigg]^2_0$

. . . . . . . . . . $=\;\;\frac{\pi}{4}(2^4) - \frac{\pi}{4}(0^4) \;\;=\;\;\frac{\pi}{4}(16) \;\;=\;\;\boxed{4\pi}$

**Legendsn3verdie** · September 12th 2008, 01:21 PM

Originally Posted by Soroban

Hello, Legendsn3verdie!

Sorry, I don't understand any of your integrals . . .

Code:

        |
        |             *
       2+ - - - * - - -
        |:::*
        |:*
        |*
        |
    - - * - - - - - - - - -
        |

Formula: . $V \;=\; \pi\int^b_a \!\!x^2\,dy$

We have: . $V \;\;=\;\;\pi\int^2_0\!\! \left(y^{\frac{3}{2}}\right)^2\,dy \;\;=\;\; \pi\int^2_0\!\! y^3\,dy \;\;=\;\;\frac{\pi}{4}y^4\,\bigg]^2_0$

. . . . . . . . . . $=\;\;\frac{\pi}{4}(2^4) - \frac{\pi}{4}(0^4) \;\;=\;\;\frac{\pi}{4}(16) \;\;=\;\;\boxed{4\pi}$

ah ha! ty sir!

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