1. ## [SOLVED] Volume

A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.

2. Originally Posted by chukie
A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.
$A=\pi\int_a^{b}A(x)dx$

where $A(x)$ is to be determined by finding in this case what the radius would be in $A=\frac{1}{2}\pi{r^2}$

3. Originally Posted by chukie
A solid has, as its base, the circular region in the xy-plane bounded by the graph of x^2+y^2=4. FInd the volume of the solid if every cross section by a plane perpendicular to the x-axis is a quarter circle with one of its radii in the base.

I really can't picture this. I tried to draw it out, but I'm unable to get the right answer.
The area is just $\int_a^b A(x)\,dx$, where $A(x)$ is the area of the cross section at $x$. Since the cross section is a quarter circle, its area is $\frac14\pi r^2$. We can find $r$ to be the length of the chord in the base perpendicular to the $x$-axis at any given value of $x$, so $r = 2\sqrt{4-x^2}$. Since the base is a circle of radius 2 centered at the origin, the cross sections are taken for $x\in[-2,2]$. The total area is then

$\int_{-2}^2\frac14\pi\left(2\sqrt{4 - x^2}\right)^2\,dx$

$=\pi\int_{-2}^2\left(4 - x^2\right)\,dx$

4. chukie, I made up a quick graph of this since you said that you had difficulty visualizing it.

Note that we have a circular base, and along the left half of it lies the "center" of our quarter-circles (what I mean to say is, the center of the circles that the quarter-circles belong to), and the radius of these circles extend from one side of the base to the other.

Originally Posted by Mathstud28
$A=\pi\int_a^{b}A(x)dx$

where $A(x)$ is to be determined by finding in this case what the radius would be in $A=\frac{1}{2}\pi{r^2}$
Almost. We are dealing with cross sections, not solids of revolution, so you do not need the $\pi$ in front of the integral. Also, the cross sections are quarter circles, so the formula we want is $A = \frac14\pi r^2$. Not sure how I missed this, since it says you posted before me.