problem: lets take the curve y = 2sin(x) and create a solid by constructing semi-circular cross...
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Problem: Lets take the curve y = 2sin(x) and create a solid by constructing semi-circular cross sections perpendicular to the x axis (the XY plane) whose diameters extend from the x-axis to the curve. You can think of an individual cross section as a semicircular fin sticking up out of the XY plane. What would be the volume of this solid?
0
)sin(2 xy
0
360 View
0
0
)sin(2 x
2
2rArea circlesemi
)sin(2
)sin(2
2x
xdiameterr
2
)][sin( 2xArea circlesemi
0
2
)][sin( 2xArea circlesemi
dx 2sin(x)
2sin(x)
dxx
Volumefin
circularsemi 2
)][sin( 2
0
In the diagram above we only have 7 semi-circular cross sections.
0
• But the 3D object we are designing has an infinite number of these cross sections,so that the figure will look like the one above
The sum of the volumes of an infinite number of semi-circular cross sectional fins with diameter 2sin(x) and thickness dx, on the interval
0
2
2
)][sin(dx
x],0[
This sum can be representedby the following integral:
will give us
the volume of the final 3D object.
Why?
Recall that the volume of each cross sectional disk was:
dx 2sin(x)
dxx
Volumedisk
circularsemi 2
)][sin( 2
If we want to sum up an infinite number of these disksWe simply integrate this volume over the desired interval:
0
2
2
)][sin(dx
x
Steps for Finding Volume of Cross Sections
1) Find an expression for the area of a single generic cross section. Let’s call that expression A(x).
2) Multiply A(x) by dx (the thickness of the cross-sectional disk) to get the volume of the cross sectional disk.
3) Integrate A(x)dx over the interval in question: b
adxxA )(
Another Example (taken from Problem 1 of the 2000 AP):
Let R be the region in the first quadrant enclosed by the graphs of :
)cos(12
xyandey x and the y axis (as shown).
2xey
)cos(1 xy
The region R is the base of a solid. For this solid, each crosssection perpendicular to the x-axis is a square. Find the volume of this solid.
Let’s try to picture what this solid looks like:
R
2) Now let’s isolate our region R and lay it flaton a 3 dimensional coordinate system
3D Solid
1) Here is our region R
Flatten 3D
3) Now let’s add square cross sections perpendicular to the X axis
4) Here is what our final solid looks like
So what would be the volume of this solid?
1) Let’s take one square cross section and find its area:
2)))cos(1()(()(2
xexA x
))cos(1()(2
xe x
dx
Since our 2 curves intersect at 941.x
461.0
)))cos(1()(()(947.
0
2947.
0
2
dxxedxxAV x