# Thread: volumes of solids by revolution

1. ## volumes of solids by revolution

find the volume of the solid of revolution generated by revolving about the y-axis the region bounded by the curve $y = \sqrt{x}$, the x-axis and the line x = 4;

Just checking out if its:
$\pi \int_{0}^{2} (4)^2 - (4 - y^2)^2 dy$ ?

2. Originally Posted by ^_^Engineer_Adam^_^
find the volume of the solid of revolution generated by revolving about the y-axis the region bounded by the curve $y = \sqrt{x}$, the x-axis and the line x = 4;

Just checking out if its:
$\pi \int_{0}^{2} (4)^2 - (4 - y^2)^2 dy$ ?
why do you have $\left( 4 - y^2 \right)^2$? it is the out radius, which is 4. minus the inner radius which is $y^2$

your formula should be:

$\pi \int_{0}^{2} \left[4^2 - \left( y^2 \right)^2 \right]~dy$

(remember, the $\pi$ should be applied to both radii)

...you know, I never noticed you were from the Philippines until now...

3. Originally Posted by ^_^Engineer_Adam^_^
find the volume of the solid of revolution generated by revolving about the y-axis the region bounded by the curve $y = \sqrt{x}$, the x-axis and the line x = 4;

Just checking out if its:
$\pi \int_{0}^{2} (4)^2 - (4 - y^2)^2 dy$ ?
outer radius = 4
inner radius = x

dV = pi[4^2 -x^2]dy

y = sqrt(x)
y^2 = x
so,
x^2 = y^4

Therefore,
dV = pi[4^2 -y^4]dy