Question

Let D1 be the region in the XY-plane enclosed by the triangle with vertices (0, 0),(1, 1),(2, 0). Let D2 be the semi-circular region of radius 1 with the center at (1, 0) lying below the x-axis. Let D = D1 ∪ D2.

a) Argue whether D is an elementary region.

b) Compute the area of D.

c) Without using Green’s theorem, evaluate $\intop$_C(xydx + x²dy), where C denotes the boundary of D oriented clockwise.

d) Use Green’s theorem to evaluate the line integral given in part c).

Answer 1

(a.)

Given: D₁: Region in the XY plane enclosed by the triangle with vertices (0,0)(1,1)(2,0)

D₂: Semicircular region with radius 1, with center at (1,0) lying below the x-axis

D=D₁ ∪ D₂

Plotting and graphing the required points and curves, the graph of the boundary of D is given as follows:

Note that the part of the boundary of D above x-axis (marked in red) is nothing but the graph of the function,

$y=f_1(x)=1−|x−1|, |x−1|<1$

Similarly, the part of the boundary of D

D below x-axis (marked in black) is nothing but the graph of the function,

$y=f_2(x)=−\sqrt{1−(x−1)^2}$

It follows that the region D can be bounded between the two functions y=f₁(x) and y=f₂(x) and thus, by definition, is a y-simple region.

Note that, it is not possible to define functions x=g₁(y) and x=g₂y so that the region D is bounded between them. Thus, D is not a x-simple region.

A region is said to be elementary if it is y-simple or x-simple.

So, D is y-simple but not x-simple and thus, by definition, is elementary.

(b.)

According to the given data, D=D₁∪D₂, where D₁ is the region in the XY plane enclosed by the triangle with vertices (0,0) (1,1)(2,0)2,0 and D₂ is the semicircular region with radius 1, with center at (1,0) lying below the x-axis.

The graph of D₁ is given as,

Similarly, the graph of D₂ is given as,

It is clear from these graphs that the regions D₁ and D₂ are non-intersecting. So, the area of 

D=D1∪D2 is nothing but the sum of areas of D1 and D2

 

Now, the area of D1 is the area of the triangle with vertices at (0,0), (1,1), (2,0)

Let A:(0,0), B:(1,1) C:(2,0)

Let BC=a, AC=b and AB=c

Using distance formula, we have,

$BC=\sqrt{z(1-2)^2+(1-0)^2}=\sqrt{2}$

$AC=\sqrt{(0-2)^2+(0-0)^2}=2$

$AB=\sqrt{(0-1)^2+(0-1)^2}=\sqrt{2}$

so

$a=\sqrt 2,\space b=2,\space c=\sqrt 2$

The semi-perimeter is given as,

$s=\frac{a+b+c}{2}=\frac{\sqrt 2+2+\sqrt2}{2}=1+\sqrt2$

now,

$s-a=1+\sqrt2-\sqrt2=1\\s-b=1+\sqrt2-2=-1+\sqrt2\\s-c=1+\sqrt2-\sqrt2=1$

Now, the area of the triangle is given as,

$A=\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{(1+\sqrt2)(1)(1\sqrt2-1)\\=1}$

So, it follows that the area of the region D1 is 1 sq. units

 

Now, the area of the region D2 is the area of the semicircular region with radius 1

1 and center at (1,0) lying below the x-axis. That is, the area of D2

is the area of a semi-circle with radius 1.

So, the required area is,

$A=\frac{\pi r^2}{2}=\frac{\pi (1)^2}{2}=\frac{\pi}{2}$

So, it follows that the area of the region D2 is  $\frac{\pi}{2}$ sq. units

Finally, the area of D $\frac{\pi}{2}$ +1 square units.

(c)

The given integral is$∫_Cxydx+x^2dy,$ where C denotes the boundary of D oriented clockwise.

The graph of C is given as,

It is given that C is oriented clockwise. So, the path of C can be divided into the following paths:

(i.) C1: The straight line from (0,0) to (1,1)

(ii.) C2: The straight line from (1,1)to (2,0)

(iii.) C3: The semicircular arc from (2,0) to (0,0)

Path C1:

Now, using two point formula, the equation of the straight line from (0,0) to (1,1) is given as,

$\frac{y-0}{x-0}=\frac{1-0}{1-0}\implies y=x$

So, the path C1 has equation y=x Then, here dy=dx.

Path C2:

Now, using two point formula, the equation of the straight line from (1,1)

 to (2,0) is given as,

$\frac{y-0}{x2}=\frac{1-0}{1-2}$

⇒y=−x+2

So, the path C2 has equation y=−x+2. Then, here dy=−dx.

Path C3:

The path C3 is the semicircular arc from (2,0 to (0,0)

As shown previously, it has the equation $y=\sqrt{−1−(x−1) ^2}$ .

Then, here,

$dy=d(-\sqrt{1-(x-1)^2})$

$=-\frac{1}{2\sqrt{1-(x-1)^2}}d(1-(x-1)^2$

$=\frac{x-1}{\sqrt{1-(x-1)^2}}$

We can write,

$∫_C(xydx+x^2dy)=∫_{c1}(xydx+x^2dy)+∫_{C2}(xydx+x^2dy)+∫_{C3}(xydx+x^2dy)$

now,

$∫_{C1}(xydx+x^2dy)=∫_1^0(x^2dx+x^2dx)\\=∫_1^02x^2dx\\=\frac{2}{3}$

Similarly,

$∫_{C2}(xydx+x^2dy)=∫_2^1(x(−x+2)dx+x^2(−dx))\\=-2\int_1^2(-2x^2dx+2xdx-x^2dx)\\=-2|\frac{x^3}{3}-\frac{x^2}{2}|_1^2\\=-\frac{5}{3}$

$Lastly, ∫_{C3}(xydx+x^2dy)=∫_0^2(x(−\sqrt{1−(x−1)^2})dx+x^2(\frac{x-1}{\sqrt{1-(x-1)^2}})dx))\\=\int_2^0(-x\sqrt{1-(x1)^2}dx+\frac{x^3-x^2}{\sqrt{1-(x-1)^2}}dx\\=\int_2^0(\frac{2x^3-3x^2}{\sqrt{1-(x1)^2}}dx$

$Let x−1=u. Then, x=u+1\space and\space dx=du. Also, x=2⇒u=1\space and\space x=0⇒u=−1$

$∫-{C3}(xydx+x^2dy)=∫^{−1}_1\frac{(2(u+1)^3−3(u+1)^2}{\sqrt{1−u^2}}du)$

$Now, let \space u=sin\space v. Then, v=sin^{−1}u\space and\space du=cos\space vdv.$

$So, ∫(\frac{2(u+1)^3−3(u+1)^2}{\sqrt{1−u^2}}du)=−\frac{3}{2}cos\space v+\frac{1}{6}cos\space 3v+\frac{1}{2}v−\frac{3}{4}sin\space 2v$

Now, u=sin v

Then,

$cos\space v=\sqrt{1−sin^2 v}=\sqrt{1−u^2}\\ cos3\space v=4cos^3v−3cos\space v\\=4(1−u^2)^{\frac{3}{2}}−3(1−u^2)^{\frac{1}{2}}\\ sin\space 2v=2sinvcosv=2u(1−u^2)^\frac{1}{2}$

Using these values, we have,

$∫^1_{−1}(\frac{2(u+1)^3−3(u+1)^2}{\sqrt{1−u^2}}du)=\frac{\pi}{2}$

$So, ∫_C(xydx+x^2dy)=\frac{π}{2}−1$

This gives the value of the required integral.

d.

m=xy ,N=x²

$\int_cxydx4x^2dy=\iint_D(2x-x)dxdy\\\iint_{D1}xdxdy-\iint_{D2}xdxdy\\\iint_{D2}xdxdy=\int_0^{\pi}\intop_0^{1}(1+rcosQ)r d0dQ\\\pi(\frac{1}{2}-0)+0(\frac{1}{3}-0)\\=\frac{\pi}{2}+0$

Answer:

(a.) The region D is an elementary region.

(b.) The area of is $\frac{\pi}{2}+1$ square units.

(c.) ∫$_C (xydx+x ^2 dy) = \frac{π}{ 2} −1$

(d)$-\frac{\pi}{2}+1$