Volume of a solid $y = 5\sqrt{ 9 − x^2}$ , $y = 0$, $x = 0$, $x = 1$; about the $x$-axis Find the volume $V$ of the solid obtained by rotating the region bounded by the given curves about the specified line.
$$y = 5 \sqrt {9 − x^2}$$
$$y = 0, x = 0, x = 1$$
about the $x-$axis 
 A: *

*Draw out the region.  The three lines $y=0, x=0, x=1$ are straight lines, and the last one $y=5\sqrt{9-x^2}$ is a curve.  (Where does the last curve intersect the other lines you have?)

*Imagine sticking a long pin through the bottom part of the region (like you're barbecuing), along the $x$ axis.  Now, walk over to some point on the negative $x$ axis, like $(-2,0)$, and look in the direction of the positive $x$ axis.

*Imagine also that the part of the region that intersects $x=0$ is red, and that you can see it from where you're standing.

*Now, spin the pin around.  See how the red line segment traces out a circle?  The cross section of all of the parts going along the $x$ axis are circles, but with different sizes.


To find the volume, then, you add up the volumes of all of the thin circular pancakes (of thickness $dx$) between $0..1$.  Let's say the area of the circle at position $x$ is $A(x)$:
$$V = \int_0^1 A(x)dx.$$
Since the area of a circle of radius $r$ is $\pi r^2$, we're just about done, because we know that the radius of the slice at position $x$ is $5\sqrt{9-x^2}$:
$$V = \int_0^1 \pi \left(5\sqrt{9-x^2}\right)^2 dx.$$
Can you take it from here?
A: If we have $$y=5\sqrt{9-x^2}\quad,\quad 0\le x\le1$$ then 
$$V=\pi\int_{0}^{1}(5\sqrt{9-x^2})^2dx=25\pi\int_{0}^{1}(9-x^2)dx$$
