Answer to Question #221505 in Calculus for Gracey

Question

(Sections 13.1–13.6) Let R be the region in the sketch below. MAT2615/001 (a) Describe the region R as a union of two Type 1 regions. (Use set–builder notation.) Hints: • Read the description of a Type 1 Region on p.18 of Guide 3 and study Fig. 13.9 carefully. • Shade the region R by means of vertical lines and highlight the curves which form lower and upper boundaries of R. Write the equations of these curves in the form y = g(x). • The lower boundary of R is formed by two different curves. (3) (b) Describe the region R as a union of two Type 2 regions. (Use set–builder notation.)

Answer 1

Hint : Since the customer did not indicate any picture that is necessary, since in the task itself they ask to describe the integration area, so I will assume that it is necessary to solve the task that is indicated on the photo I provided.

Preparatory work: we will find the coordinates of the intersection point of the graphs $y=x$ and $x^2+y^2=4$ , for this we will solve the system

$$\left\{\begin{array}{l} x^2+y^2=4\\[0.3cm] y=x \end{array}\right.\longrightarrow x^2+x^2=4\longrightarrow x^2=2\\[0.3cm] |x|=\sqrt{2}\longrightarrow\boxed{x=y=\sqrt{2}}$$

QUESTION A

We will draw any straight line $x=b$ so that it is within our region. We see that such a line crosses the borders of the region twice. lower point corresponds to the lower boundary of the region, and the top - the top.

The upper boundary :

$$\left\{\begin{array}{l} x^2+y^2=4\\[0.3cm] y\ge 0 \end{array}\right.\longrightarrow\boxed{y=f(x)=\sqrt{4-x^2}}$$

The lower boundary :

$$y=g(x)=\left\{\begin{array}{l} y=1,\,\,\,0\le x\le1\\[0.3cm] y=x,\,\,\,1\le x\le \sqrt{2} \end{array}\right.$$

ANSWER

$$y=f(x)=\sqrt{4-x^2}-\text{the upper boundary}\\[0.3cm] y=g(x)=\left\{\begin{array}{l} y=1,\,\,\,0\le x\le1\\[0.3cm] y=x,\,\,\,1\le x\le \sqrt{2} \end{array}\right.-\text{the lower boundary}$$

QUESTION B

We will draw any straight line $y=b$ so that it is within our region. We see that such a line crosses the borders of the region twice. the left one corresponds to the left border of the region, and the right one corresponds to the right one.

The left boundaty :

$$\boxed{x=0}$$

The right boundary :

$$x=g(y)=\left\{\begin{array}{l} x=y,\,\,\,1\le y\le\sqrt{2}\\[0.3cm] x=\sqrt{4-y^2},\,\,\,\sqrt{2}\le y\le 2 \end{array}\right.$$

ANSWER

$$x=0-\text{the left boundaty}\\[0.3cm] x=g(y)=\left\{\begin{array}{l} x=y,\,\,\,1\le y\le\sqrt{2}\\[0.3cm] x=\sqrt{4-y^2},\,\,\,\sqrt{2}\le y\le 2 \end{array}\right.-\text{the right boundary}$$

QUESTION C

Let's change to the polar coordinate system

$$\left\{\begin{array}{l} x=r\cdot\cos\theta\\[0.3cm] y=r\cdot\sin\theta \end{array}\right.$$

Now we write each of the functions described in QUESTION A in the form of polar functions:

$$x=0\longrightarrow r\cdot\cos\theta=0\longrightarrow\cos\theta=0\\[0.3cm] \boxed{\theta\le\frac{\pi}{2}}\\[0.3cm] y=x\longrightarrow r\cdot\sin\theta=r\cdot\cos\theta\longrightarrow\tan\theta=1\\[0.3cm] \boxed{\theta\ge\frac{\pi}{4}}\\[0.3cm] x^2+y^2=4\longrightarrow \left(r\cdot\cos\theta\right)^2+\left(r\cdot\sin\theta\right)^2=4\\[0.3cm] r^2=4\longrightarrow\boxed{r\le2}\\[0.3cm] y=1\longrightarrow r\cdot\sin\theta=1\longrightarrow\boxed{r\ge\frac{1}{\sin\theta}}$$

ANSWER

$$R :\left\{\begin{array}{l} \displaystyle\frac{\pi}{4}\le\theta\le\displaystyle\frac{\pi}{2}\\[0.3cm] \displaystyle\frac{1}{\sin\theta}\le r\le2 \end{array}\right.$$