question: How integrals can calculate areas?
comment: > how could it ever be accurate? As you add more rectangles the closer you get to the true answer (because each rectangle will better approximate the curve). The more rectangles you add the closer the "error" is to zero. Once you have an infinite number of them the error *is* zero. If you know what limits are it's quite literally the same thing. If every step has you get closer to reality, and you take an infinite number of steps, you'll get infinitely close to reality, which means you've arrived at the correct answer.
passages: Section::::History.
  
  Section::::History.:Pre-calculus integration.
  
  The first documented systematic technique capable of determining integrals is the method of exhaustion of the ancient Greek astronomer Eudoxus ("ca." 370 BC), which sought to find areas and volumes by breaking them up into an infinite number of divisions for which the area or volume was known. This method was further developed and employed by Archimedes in the 3rd century BC and used to calculate areas for parabolas and an approximation to the area of a circle.
   Let us notice that we defined the surface integral by using a parametrization of the surface "S". We know that a given surface might have several parametrizations. For example, if we move the locations of the North Pole and the South Pole on a sphere, the latitude and longitude change for all the points on the sphere. A natural question is then whether the definition of the surface integral depends on the chosen parametrization. For integrals of scalar fields, the answer to this question is simple, the value of the surface integral will be the same no matter what parametrization one uses.
   For an object moving along a path in a vector field such as an electric field or gravitational field, the total work done by the field on the object is obtained by summing up the differential work done in moving from to . This gives the line integral
  
  Section::::Extensions.:Surface integrals.
   Integrals appear in many practical situations. If a swimming pool is rectangular with a flat bottom, then from its length, width, and depth we can easily determine the volume of water it can contain (to fill it), the area of its surface (to cover it), and the length of its edge (to rope it). But if it is oval with a rounded bottom, all of these quantities call for integrals. Practical approximations may suffice for such trivial examples, but precision engineering (of any discipline) requires exact and rigorous values for these elements.
   To find an explicit formula for the surface integral, we need to parameterize the surface of interest, "S", by considering a system of curvilinear coordinates on "S", like the latitude and longitude on a sphere. Let such a parameterization be x("s", "t"), where ("s", "t") varies in some region "T" in the plane. Then, the surface integral is given by
   To start off, consider the curve between and with (see figure). We ask:
  
  and call this (yet unknown) area the (definite) integral of . The notation for this integral will be
   Lastly, there are surfaces which do not admit a surface normal at each point with consistent results (for example, the Möbius strip). If such a surface is split into pieces, on each piece a parametrization and corresponding surface normal is chosen, and the pieces are put back together, we will find that the normal vectors coming from different pieces cannot be reconciled. This means that at some junction between two pieces we will have normal vectors pointing in opposite directions. Such a surface is called non-orientable, and on this kind of surface, one cannot talk about integrating vector fields.
   Another issue is that sometimes surfaces do not have parametrizations which cover the whole surface. The obvious solution is then to split that surface into several pieces, calculate the surface integral on each piece, and then add them all up. This is indeed how things work, but when integrating vector fields, one needs to again be careful how to choose the normal-pointing vector for each piece of the surface, so that when the pieces are put back together, the results are consistent. For the cylinder, this means that if we decide that for the side region the normal will point out of the body, then for the top and bottom circular parts the normal must point out of the body too.
   The surface area can be calculated by integrating the length of the normal vector formula_20 to the surface over the appropriate region "D" in the parametric "uv" plane:
  
  Although this formula provides a closed expression for the surface area, for all but very special surfaces this results in a complicated double integral, which is typically evaluated using a computer algebra system or approximated numerically. Fortunately, many common surfaces form exceptions, and their areas are explicitly known. This is true for a circular cylinder, sphere, cone, torus, and a few other surfaces of revolution.
   Many problems in mathematics, physics, and engineering involve integration where an explicit formula for the integral is desired. Extensive tables of integrals have been compiled and published over the years for this purpose. With the spread of computers, many professionals, educators, and students have turned to computer algebra systems that are specifically designed to perform difficult or tedious tasks, including integration. Symbolic integration has been one of the motivations for the development of the first such systems, like Macsyma.
   For an example of applications of surface integrals, consider a vector field on a surface ; that is, for each point in , is a vector. Imagine that we have a fluid flowing through , such that determines the velocity of the fluid at . The flux is defined as the quantity of fluid flowing through in unit amount of time. To find the flux, we need to take the dot product of with the unit surface normal to at each point, which will give us a scalar field, which we integrate over the surface:
   Computations of volumes of solids of revolution can usually be done with disk integration or shell integration.
  
  Specific results which have been worked out by various techniques are collected in the list of integrals.
  
  Section::::Computation.:Symbolic.
   is defined informally as the signed area of the region in the -plane that is bounded by the graph of , the -axis and the vertical lines and . The area above the -axis adds to the total and that below the -axis subtracts from the total.
   This argument is actually a simple application of the ideas of calculus. In ancient times, the method of exhaustion was used in a similar way to find the area of the circle, and this method is now recognized as a precursor to integral calculus. Using modern methods, the area of a circle can be computed using a definite integral:
  
  Section::::Area formulas.:Area of curved shapes.:Ellipses.
  
  The formula for the area enclosed by an ellipse is related to the formula of a circle; for an ellipse with semi-major and semi-minor axes and the formula is:
  
  Section::::Area formulas.:Area of curved shapes.:Surface area.
   Section::::Direct methods.:Example.
  
  A fundamental result in complex analysis is that the contour integral of is , where the path of the contour is taken to be the unit circle traversed counterclockwise (or any positively oriented Jordan curve about 0). In the case of the unit circle there is a direct method to evaluate the integral
  
  In evaluating this integral, use the unit circle as a contour, parametrized by , with , then and
  
  which is the value of the integral.
  
  Section::::Applications of integral theorems.
   For integrals of vector fields, things are more complicated, because the surface normal is involved. It can be proven that given two parametrizations of the same surface, whose surface normals point in the same direction, one obtains the same value for the surface integral with both parametrizations. If, however, the normals for these parametrizations point in opposite directions, the value of the surface integral obtained using one parametrization is the negative of the one obtained via the other parametrization. It follows that given a surface, we do not need to stick to any unique parametrization; but, when integrating vector fields, we do need to decide in advance which direction the normal will point to and then choose any parametrization consistent with that direction.
   where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of x("s", "t"), and is known as the surface element. The surface integral can also be expressed in the equivalent form
  
  where "g" is the determinant of the first fundamental form of the surface mapping x("s", "t").
  
  For example, if we want to find the surface area of the graph of some scalar function, say formula_3, we have
  
  where formula_5. So that formula_6, and formula_7. So,
   Consider a vector field v on "S", that is, for each x in "S", v(x) is a vector.
  
  The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. This applies for example in the expression of the electric field at some fixed point due to an electrically charged surface, or the gravity at some fixed point due to a sheet of material.
   The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighting distinguishes the line integral from simpler integrals defined on intervals. Many simple formulas in physics have natural continuous analogs in terms of line integrals; for example, the fact that work is equal to force, , multiplied by displacement, , may be expressed (in terms of vector quantities) as:
   The area of the blue region is
  
  Similarly, the area of the red region is
  
  The total area "A" + "A" is equal to the area of the bigger rectangle, "x""y", minus the area of the smaller one, "x""y":
  
  Or, in terms of "t",
  
  Or, in terms of indefinite integrals, this can be written as
  
  Rearranging:
  
  Thus integration by parts may be thought of as deriving the area of the blue region from the area of rectangles and that of the red region.
   A similar method was independently developed in China around the 3rd century AD by Liu Hui, who used it to find the area of the circle. This method was later used in the 5th century by Chinese father-and-son mathematicians Zu Chongzhi and Zu Geng to find the volume of a sphere (; ).
   Consider, for example, the integral
  
  which has the exact answer . (In ordinary practice, the answer is not known in advance, so an important task — not explored here — is to decide when an approximation is good enough.) A “calculus book” approach divides the integration range into, say, 16 equal pieces, and computes function values.
   For example, in probability theory, integrals are used to determine the probability of some random variable falling within a certain range. Moreover, the integral under an entire probability density function must equal 1, which provides a test of whether a function with no negative values could be a density function or not.
  
  Integrals can be used for computing the area of a two-dimensional region that has a curved boundary, as well as computing the volume of a three-dimensional object that has a curved boundary. The area of a two-dimensional region can be calculated using the aforementioned definite integral.
   Section::::Spatial obstructions.:Dust.
  
  Consider a space filled with small impenetrable particles, or "dust". Denote the fraction of space excluding a monomer end point by formula_30 so its values range: formula_31.
  
  Constructing a Taylor expansion for formula_26, one can arrive at the new governing differential equation:
  
  For which the corresponding path integral is:
  
  Section::::Spatial obstructions.:Walls.
  
  To model a perfect rigid wall, simply set formula_35 for all regions in space out of reach of the polymer due to the wall contour.
   BULLET::::- If the integral can be shown to be zero, or if the real-valued integral that is sought is improper, then if we demonstrate that the integral as described above tends to 0, the integral along will tend to the integral around the contour .
  
  BULLET::::- conclusion
  
  BULLET::::- If we can show the above step, then we can directly calculate , the real-valued integral.
  
  Section::::Applications of integral theorems.:Example 1.
  
  Consider the integral
  
  To evaluate this integral, we look at the complex-valued function
  