# Answer in Calculus for sai #261999

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 + x^{2}dy), where C denotes the boundary of D oriented clockwise.

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

(a.)

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

D_{2}: Semicircular region with radius 1, with center at (1,0) lying below the x-axis

D=D_{1} ∪ D_{2}

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)=1u2212|xu22121|, |xu22121|<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)=u2212sqrt{1u2212(xu22121)^2}”

It follows that the region D can be bounded between the two functions y=f_{1}(x) and y=f_{2}(x) and thus, by definition, is a y-simple region.

Note that, it is not possible to define functions x=g_{1}(y) and x=g_{2}y 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=D_{1}∪D_{2}, where D_{1} is the region in the XY plane enclosed by the triangle with vertices (0,0) (1,1)(2,0)2,0 and D_{2} is the semicircular region with radius 1, with center at (1,0) lying below the x-axis.

The graph of D_{1} is given as,

Similarly, the graph of D_{2} is given as,

It is clear from these graphs that the regions D_{1} and D_{2 }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)(1sqrt2-1)\=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“u222b_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{u22121u2212(xu22121)nn^2}” .

Then, here,

“dy=d(-sqrt{1-(x-1)^2})”

“=-frac{1}{2sqrt{1-(x-1)^2}}d(1-(x-1)^2”

“=frac{x-1}{sqrt{1-(x-1)^2}}”

We can write,

“u222b_C(xydx+x^2dy)=u222b_{c1}(xydx+x^2dy)+u222b_{C2}(xydx+x^2dy)+u222b_{C3}(xydx+x^2dy)”

now,

“u222b_{C1}(xydx+x^2dy)=u222b_1^0(x^2dx+x^2dx)\=u222b_1^02x^2dx\=frac{2}{3}”

Similarly,

“u222b_{C2}(xydx+x^2dy)=u222b_2^1(x(u2212x+2)dx+x^2(u2212dx))\=-2int_1^2(-2x^2dx+2xdx-x^2dx)\=-2|frac{x^3}{3}-frac{x^2}{2}|_1^2\=-frac{5}{3}”

“Lastly,nnu222b_{C3}(xydx+x^2dy)=u222b_0^2(x(u2212sqrt{1u2212(xu22121)^2})dx+x^2(frac{x-1}{sqrt{1-(x-1)^2}})dx))\=int_2^0(-xsqrt{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 xu22121=u. Then, x=u+1space andspace dx=du. Also, x=2u21d2u=1space andspace x=0u21d2u=u22121”

“u222b-{C3}(xydx+x^2dy)=u222b^{u22121}_1frac{(2(u+1)^3u22123(u+1)^2}{sqrt{1u2212u^2}}du)”

“Now, let space u=sinspace v. Then, v=sin^{u22121}uspace andspace du=cosspace vdv.”

“So, u222b(frac{2(u+1)^3u22123(u+1)^2}{sqrt{1u2212u^2}}du)=u2212frac{3}{2}cosspace v+frac{1}{6}cosspace 3v+frac{1}{2}vu2212frac{3}{4}sinspace 2v”

Now, u=sin v

Then,

“cosspace v=sqrt{1u2212sin^2 v}=sqrt{1u2212u^2}\ncos3space v=4cos^3vu22123cosspace v\=4(1u2212u^2)^{frac{3}{2}}u22123(1u2212u^2)^{frac{1}{2}}\ nsinspace 2v=2sinvcosv=2u(1u2212u^2)^frac{1}{2}”

Using these values, we have,

“u222b^1_{u22121}(frac{2(u+1)^3u22123(u+1)^2}{sqrt{1u2212u^2}}du)=frac{pi}{2}”

“So, u222b_C(xydx+x^2dy)=frac{u03c0}{2}u22121”

This gives the value of the required integral.

d.

m=xy ,N=x^{2}

“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.) ∫“_Cnn(xydx+xnn^2nndy)nn=nnfrac{u03c0}{nn2}nnnnu22121”

(d)“-frac{pi}{2}+1”