A right triangle has side lengths $3$, $4$, and $5.$ A circle is then drawn with each side as a diameter. Find the area of the region contained in all three circles.
Let $O_1$, $O_2$, $O_3$ be the centers of the circles with diameters 3, 4, and 5, respectively.
[asy]
unitsize(0.4 cm);
pair A, B, C, O1, O2, O3;
A = (0,0);
C = (5,0);
O3 = extension(A, A + (C - A)/(C - A).rotate(90), C, C + (A - C)/(A - C).rotate(90));
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;
draw(A--B--C--cycle);
draw(arc(O1,3/2,0,180));
draw(arc(O2,4/2,0,180));
draw(arc(O3,5/2,0,180));
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);
dot("$O_1$", O1, N);
dot("$O_2$", O2, SW);
dot("$O_3$", O3, dir(270));
[/asy]
We rotate $\triangle ABC$ about $O_3$ by 180 degrees, to form a new triangle with side lengths 3, 4, and 5.
[asy]
unitsize(0.4 cm);
pair A, B, C, O1, O2, O3;
A = (0,0);
C = (5,0);
O3 = (0,0);
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;
draw(A--B--C--cycle);
draw(O1--O3);
draw(O2--O3);
label("$A$", A, W);
label("$B$", B, N);
label("$C$", C, E);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);
dot("$O_1$", O1, SB);
dot("$O_2$", O2, dir(120));
dot("$O_3$", O3, S);
dot("$(0,3)$", (0,3), N);
[/asy]
The centers of the circles with diameters 4 and 5 are $O_2 = (0,0)$ and $O_3 = (0,3)$, respectively. Rotating the triangle about $O_3$ also rotates the lines $O_2 O_3$ and $O_1 O_3,$ so the centers of the circles with diameters 5 and 3 are mapped to points $(0,-4)$ and $(0,6)$, respectively. The equation of the circle with diameter 5 is then
\[(x - 0)^2 + (y - 3)^2 = \left( \frac{5}{2} \right)^2 = \frac{25}{4},\]the equation of the circle with diameter 4 is
\[(x - 0)^2 + (y - 4)^2 = 2^2 = 4,\]and the equation of the circle with diameter 3 is
\[(x - 0)^2 + (y - 6)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}.\]We take the horizontal line $y = k,$ for $-4 \le k \le 3,$ to slice the region that we want.
[asy]
unitsize(0.4 cm);
int i;
pair A, B, C, O1, O2, O3;
real k;
A = (0,0);
C = (5,0);
O3 = (0,0);
B = intersectionpoint(arc(O3,5),arc(C,3));
O1 = (A + B)/2;
O2 = (C + B)/2;
k = 1;
draw(A--B--C--cycle);
draw(O1--O3);
draw(O2--O3);
draw(arc(O1,3/2,0,180));
draw(arc(O2,4/2,0,180));
draw(arc(O3,5/2,0,180));
label("$A$", A, SW);
label("$B$", B, dir(190));
label("$C$", C, SE);
label("$3$", (A + B)/2, NW);
label("$4$", (B + C)/2, NE);
label("$5$", (A + C)/2, S);
dot("$O_1$", O1, N);
dot("$O_2$", O2, SW);
dot("$O_3$", O3, S);
dot("$(0,3)$", (0,3), N);
draw((0,-4)--(0,6),dashed);
draw((0,3)--(intersectionpoint(arc(O1,3/2,0,90),arc(O3,5/2,0,90)))--(intersectionpoint(arc(O2,4/2,0,90),arc(O3,5/2,0,90)))--cycle);
draw((intersectionpoint(arc(O2,4/2,90,180),arc(O3,5/2,90,180)))--(intersectionpoint(arc(O1,3/2,90,180),arc(O3,5/2,90,180)))--(0,3)--cycle);
draw((0,3)--(intersectionpoint(arc(O1,3/2,180,270),arc(O3,5/2,180,270)))--(intersectionpoint(arc(O2,4/2,180,270),arc(O3,5/2,180,270)))--cycle);
draw((intersectionpoint(arc(O2,4/2,270,360),arc(O3,5/2,270,360)))--(intersectionpoint(arc(O1,3/2,270,360),arc(O3,5/2,270,360)))--(0,3)--cycle);
label("$k = -4$", ((0,-4) + (0,-4) + intersectionpoint(arc(O2,4/2,90,180),arc(O3,5/2,90,180)))/3, N);
label("$k = 0$", ((0,0) + (0,0) + intersectionpoint(arc(O2,4/2,0,90),arc(O3,5/2,0,90)))/3, N);
label("$k = 3$", (3/2,(0,3) + (0,3))/2, N);
[/asy]
The solution is the sum of the areas of the three shaded regions. The area of each shaded region is a quarter-circle minus a triangle.
The area of the lowest shaded region is
\[\frac{1}{4} \cdot \pi \cdot \left( \frac{5}{2} \right)^2 - \frac{1}{2} \cdot 4 \cdot 4 = \frac{25 \pi}{8} - 8.\]The area of the middle shaded region is
\[\frac{1}{4} \cdot \pi \cdot 2^2 - \frac{1}{2} \cdot 3 \cdot 4 = \frac{4 \pi}{4} - 6.\]The area of the top shaded region is
\[\frac{1}{4} \cdot \pi \cdot \left( \frac{3}{2} \right)^2 - \frac{1}{2} \cdot 3 \cdot 3 = \frac{9 \pi}{16} - \frac{9}{2}.\]Hence, the area that is contained in all three circles is
\[\boxed{\frac{25 \pi}{8} + \frac{4 \pi}{4} + \frac{9 \pi}{16} - 8 - 6 - \frac{9}{2}}.\]
To find the area of the region contained in all three circles, we need to find the area of the intersection of the three circles.
Since each side of the right triangle is a diameter, the radius of each circle is half the length of the corresponding side. Thus, the radii of the circles are $3/2$, $4/2$, and $5/2$.
The intersection of the three circles forms an equilateral triangle. To find the area of an equilateral triangle, we need to know the length of its side. Let $s$ be the side length of the equilateral triangle.
In a right triangle with side lengths $3$, $4$, and $5$, the side opposite the right angle has length $3$ and is the height of the equilateral triangle. In an equilateral triangle, the height is the perpendicular drawn to the base and is also the median and the angle bisector.
Since the equilateral triangle is formed by the intersection of the circles with diameters $3$, $4$, and $5$, the length of the side $s$ is $3$.
The area $A$ of an equilateral triangle with side length $s$ is given by the formula $A = \frac{\sqrt{3}}{4} s^2$.
Plugging in $s = 3$ into the formula, we have $A = \frac{\sqrt{3}}{4} (3^2) = \frac{9 \sqrt{3}}{4}$.
Therefore, the area of the region contained in all three circles is $\frac{9 \sqrt{3}}{4}$ square units.