1. ## shell method

Find the volume of the solid formed by revolving the region bounded by the graph of f(x)=4-x^2 and the x-axis about the line x=3 using the shell method.

2. $2\pi \int_{0}^{4} rhdy$

$r = 3 - y$

$h = \sqrt {4 - y}$

$2\pi \int_{0}^{4} (3 - y)\sqrt {4 - y}dy$

u = 4 - y
dy = -dy

$-\int (u + 1)\sqrt udu = -\int (u^{3/2}du + \sqrt udu)$

3. Originally Posted by derfleurer
$2\pi \int_{0}^{4} rhdy$

$r = 3 - y$

$h = \sqrt {4 - y}$

$2\pi \int_{0}^{4} (3 - y)\sqrt {4 - y}dy$

u = 4 - y
dy = -dy

$-\int (u + 1)\sqrt udu = -\int (u^{3/2}du + \sqrt udu)$
how can u replace 3 in 3-y?