# Volume of bounded region

• Feb 17th 2010, 05:49 PM
bfpri
Volume of bounded region
Let R denote the region enclosed by the y-axis, the line y = 1, and the curve
y = sqrt(x). Compute the volume of the solid whose base is R and whose cross sections perpendicular to the y-axis are semicircles.

I don't get it.. i did integral(pi*(y^2)^2)dy) from y=0 to y=1. i got pi/10
The answer is supposed to be pi/40.
• Feb 17th 2010, 06:12 PM
Quote:

Originally Posted by bfpri
Let R denote the region enclosed by the y-axis, the line y = 1, and the curve
y = sqrt(x). Compute the volume of the solid whose base is R and whose cross sections perpendicular to the y-axis are semicircles.

I don't get it.. i did integral(pi*(y^2)^2)dy) from y=0 to y=1. i got pi/10
The answer is supposed to be pi/40.

Hi bfpri,

$\displaystyle y=\sqrt{x}$

$\displaystyle x=y^2$

and all of these semicircles have diameter x, therefore radius (0.5)x.
Their areas are half the area of a circle of the same radius.
Therefore,

$\displaystyle \frac{1}{2}\int_{0}^1{\pi}r^2dy=\frac{\pi}{2}\int_ {0}^1\left(\frac{y^2}{2}\right)^2dy=\frac{\pi}{2}\ int_{0}^1\frac{y^4}{4}dy$$\displaystyle =\frac{\pi}{8}\frac{1^5}{5}=\frac{\pi}{40}$
• Feb 17th 2010, 06:15 PM
bfpri
thanks!