# 微积分作业代写calclulus代考| Drawing or visualizing surfaces in R3

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## 微积分作业代写calclulus代考|Example 1.6:

We shall consider here the graph of the function
$$f(x, y)=\frac{4 x}{1+x^{2}+y^{2}} \quad \text { for } \quad(x, y) \in \mathbb{R}^{2}$$
This function features in an exercise in a later chapter. For now we are just interested in determining the form taken by the function’s graph,
$$G=\left{(x, y, z):(x, y) \in \mathbb{R}^{2}, z=f(x, y)\right} .$$
In the steps that follow we will in essence dissect the function, and with the pieces we obtain we will build up a picture of the graph.

Step 1: The first thing to note is the domain of definition. What you would be looking for are the limits on the independent variables as well as possible points where the function is not defined. In our case, the function is defined everywhere so the domain is the entire $x y$-plane.

Step 2: The second thing to do is to look for any zeros of the function. That is, we look for intercept points in the domain at which the function takes the value zero. Here, $f=0$ when $x=0$, that is, at all points along the $y$-axis.
Step 3: We now look for any symmetry. We note that the function is odd in $x$ but even in $y$. The symmetry in $x$ means that for any fixed $y$ – which means taking a cross-section of the graph parallel to the $x$-axis – howsoever the graph appears for $x>0$, it will be inverted in the $x y$-plane for $x<0$. The symmetry in $y$ means that for any fixed $x$ (that is, a cross-section parallel to the $y$-axis) the graph will look the same on the left of $y=0$ as on the right. Note, however, that because of the oddness in $x$, the graph will sit above the $x y$-plane for $x>0$, but below the plane for $x<0$.

## 微积分作业代写calclulus代考|Example 1.7:

Consider $S=\left{(x, y, z): x^{2}+y^{2}+z^{2}=a^{2}, a>0\right}$. This is a surface in $\mathbb{R}^{3}$, Figure $1.23$; it is a surface because there exists a relation between the three variables $(x, y, z)$. They are no longer completely independent: one variable can be considered a function of the other two.

Now set $z=0$. This simplifies to the subset satisfying $x^{2}+y^{2}=a^{2}$ which is a curve (circle) in the $x y$-plane. Note that these two equations for the three variables, which is equivalent to setting two conditions on the three variables, generate a curve in $\mathbb{R}^{2}$.

A consistent interpretation is that of the intersection of two surfaces: The plane $z=0$ and the sphere $S$ giving rise to the subset of points the surfaces have in common – the circle of radius $a$ in the $x y$-plane.
Suppose that $a>2$, say, in $S$. Then setting
\begin{aligned} &z=0 \Longrightarrow x^{2}+y^{2}=a^{2} \ &z=1 \Longrightarrow x^{2}+y^{2}=a^{2}-1<a^{2} \ &z=2 \Longrightarrow x^{2}+y^{2}=a^{2}-4<a^{2}-1<a^{2} \ &z=a \Longrightarrow x^{2}+y^{2}=a^{2}-a^{2}=0 \Longleftrightarrow x=y=0 \end{aligned}
These are examples of level sets defining circles in the $x y$-plane. We will come back to discuss these in detail in Section 1.F.

## 微积分作业代写calclulus代考|Example 1.6:

F(X,和)=4X1+X2+和2 为了 (X,和)∈R2

G=\left{(x, y, z):(x, y) \in \mathbb{R}^{2}, z=f(x, y)\right} 。G=\left{(x, y, z):(x, y) \in \mathbb{R}^{2}, z=f(x, y)\right} 。