By Piskunov

schools. It comprises many labored examples that illustrate the theoretical
material and function types for fixing problems.
The first chapters "Number. Variable. functionality" and "Limit.
Continuity of a functionality" were made as brief as attainable. a few of the
questions which are often mentioned in those chapters were installed the third
and next chapters with out lack of continuity. This has made it possible
to absorb very early the fundamental inspiration of differential calculus - the
derivative - that's required within the research of technical topics. event has
shown this association of the fabric to be the easiest and so much convenient
for the student.

A huge variety of difficulties were incorporated, lots of which
illustrate the interrelationships of arithmetic and different disciplines. The problems
are particularly chosen (and in enough quantity) for every portion of the course
thus supporting the scholar to grasp the theoretical fabric. To a wide extent,
this makes using a separate publication of difficulties pointless and extends

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Extra info for Differential and Integral Calculus

Example text

Here we can expect quite a few diﬃculties, no matter which version we are choosing. In fact, the integrand is suited for the polar coordinates, while the domain B in the (x, y)-plane is best described in rectangular coordinates. By experience, such a mixture of polar and rectangular coordinates will always give computational problems. Then we notice that if we start by ﬁrst integrating after x, then we shall immediately run into troubles with this ﬁrst integral dx . (x2 + y 2 + z 2 )2 It is possible to go through with the calculations, but they are far from elementary.

Then we get by the reduction theorem that the ϕ-integral can be factored out, I = = = = π 2 −π 2 π 2 −π 2 sin2 ϕ · cos ϕ dϕ · 1 sin3 ϕ 3 2 15 cos ϕ · B(ϕ) a 0 π 2 −π 2 · z 6 dz = 2 sin2 ϕ · z · d dz 4 z d dz B(ϕ) a z z 0 dϕ 4 d dz = 0 2 · 3 a 0 z· 1 5 z dz 5 2 1 7 2 7 · a = a . 15 7 105 C. It is seen as a weak control that the result is of the form constant·a7 as mentioned in A. 5 A. Let A be an upper half sphere of radius 2a, from which we have removed a cylinder of radius a and then halved the resulting domain by the plane x + y = 0.

In this case we write the domain in the form B = {(x, y) ∈ R2 | 0 ≤ x ≤ 2, 1 − 1 x ≤ y ≤ 1}. 2 Notice that the outer variable x must always lie between two constants, 0 ≤ x ≤ 2. e. in this particular case 1 − x ≤ y ≤ 1. 2 Then write down the double integral: 2 (2) 1 xy dS = 1− 12 x 0 B 2 xy dy dx = 1 x 1− 12 x 0 y dy dx. Calculate the inner integral, 1 1− 12 y dy = x 1 2 y 2 1 = 1− 12 x 1 2 1− 1− 2 1 x 2 = 1 2 1− 1 2 x 4 = 1 1 x − x2 . 2 8 By insertion in (2), we get 2 xy dS = x 0 B = 1 1 x − x2 dx = 2 8 1 3 1 4 x − x 6 32 2 = 0 2 1 2 1 3 x − x dx 2 8 0 5 8 1 − = .