# Thread: [SOLVED] calculus volume (disk method)

1. ## [SOLVED] calculus volume (disk method)

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.

2. 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?

3. Hello, Legendsn3verdie!

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

Find the volume of: .$\displaystyle \begin{Bmatrix}x \:=\:y^{\frac{3}{2}} \\ x \:=\:0 \\ y\:=\:2 \end{Bmatrix}$ . revolved about the y-axis
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Formula: . $\displaystyle V \;=\; \pi\int^b_a \!\!x^2\,dy$

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

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

4. Originally Posted by Soroban
Hello, Legendsn3verdie!

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

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

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

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

ah ha! ty sir!