Use integral calculus to discover irregularly shaped volumes and areas!
area between two curves
This is a fairly straightforward concept, by subtracting the area under one curve by the area under another curve, you find the area between the two curves. This allows us to find the areas of organic shapes that do not have flat sides.
When f(x) is the top curve and g(x) is the bottom curve, the formula is:
When f(x) is the top curve and g(x) is the bottom curve, the formula is:
when integrating from a lower limit to an upper limit.
For example:
if f(x)= x^2 and g(x)= x
The area between them is the integral of x-x^2 with respect to x from zero to one, the points of intersection.
The antiderivative ends up being 1/2 x^2 - 1/3 x^3, and the integral is 1/6.
For example:
if f(x)= x^2 and g(x)= x
The area between them is the integral of x-x^2 with respect to x from zero to one, the points of intersection.
The antiderivative ends up being 1/2 x^2 - 1/3 x^3, and the integral is 1/6.
Volume: disk method
For volume using the disk method, think of a circle. The area of a circle is noted as pi*r^2
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when you stretch out the end of that circle, you'll get a cylinder,
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the radius becomes a surface instead of a segment and the center becomes an axis.
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we can rotate surfaces around an axis in order to find the volume of that surface. By integrating to create the surface, replacing r with f(x), the volume is:
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when you stretch out the end of that circle, you'll get a cylinder,
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the radius becomes a surface instead of a segment and the center becomes an axis.
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we can rotate surfaces around an axis in order to find the volume of that surface. By integrating to create the surface, replacing r with f(x), the volume is:
So if you have the cubic function f(x)=x^3 and need to find the volume of the function from x boundaries zero to ten when it is rotated around the x-axis, the volume would be V= pi * S (x^3)^2 dx = pi * x^7 /7 = 4487989.51 cubic units.
volume: washer method
This is a surface minus another surface rotated around a central axis. The radius squared, instead of becoming f(x)^2, becomes f(x)^2-g(x)^2. If you use functions
volume: shell method
The volume of a shell is a cylinder of thin width. Essentially, imagine the perimeter of a circle pulled up into a third dimension. If you add up a bunch of these shells according to a change in x and a height that corresponds to a function f(x), you'll end up with a volume. As a result we can use the equation for volume V=2pi*S f(x) dx.
Also, if you have a cross-sectional area f(x)= g(x)-h(x), you get to call it a "petal," which is a rather pretty name for a mathematical figure.
Also, if you have a cross-sectional area f(x)= g(x)-h(x), you get to call it a "petal," which is a rather pretty name for a mathematical figure.