# 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 + x2dy), 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: 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)=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=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 Dare 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=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”

(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”

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