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Section A: Mathematics-II Very Short Question Solutions
Q1. Define sets and Universal sets with example.
Sol. Set : It is a well-defined collection of objects. The objects of a set are called the elements or members of that set and their membership is defined by certain conditions.
For example;
- 1. The collection of all the letters of English alphabet a, b, c, d, …
- 2. The collection of all natural numbers denoted by N.
- 3. The collection of vowels in English alphabet.
Universal Set : When we are given a particular set and we consider different subsets of the given set, this given set is called universal set. It is denoted by U.
For example;
- 1. The universal set is the set of real numbers R, while considering the set of natural numbers, whole numbers, integers and rational numbers.
- 2. The set of alphabets is the universal set from which the letters of any word may be chosen to form a set.
- 3. In geometry, we discuss set of lines, triangles and circles, then universal set is the plane, in which the lines, triangles and circles lie.
Q2. Define equivalence Relation and show that the relation S ={(a, b) :a ≥ b} on the set R of real no is an equivalence relation.
Sol. A relation R on a set E is said to be an equivalence relation, if it is :
- (a) reflexive
- (b) symmetric, and
- (c) transitive.
Check for Reflexive:
It is given that R = {(a, b) : a ≥ b}
It is clear that (a, a) ∈R as a = a
Therefore, R is reflexive.
Check for Symmetric:
If a ≥ b, then b ≥ a.
This statement is true only for the case a=b.
Therefore, R is symmetric.
Check for Transitivity:
Now let (a, b) (b, c) ∈R
Then a ≥ b and b ≥ c
⇒ a ≥ c
⇒ (a, c) ∈R
Therefore, R is a transitive.
Hence, R is reflexive, transitive and symmetric. So, R is an equivalence relation.
Q3. Show that the inclusion relation ⊆ is a partial ordering on the power set of a set S.
Sol. Since A ⊆ A for any subset A ⊆ S, we conclude that this relation is reflexive.
Taking into account that A ⊆ B and B ⊆ A imply A = B, we conclude that the relation is antisymmetric.
Since A ⊆ B and B ⊆ C imply A ⊆ C,it follows that this relation is transitive.
Consequently the inclusion relation is a partial ordering on the power set of a set S. Ans.
Q4. If z = exy2, x = t cost, y = t sint compute dz/dt at t = 𝜋/2.
Sol. Given z = exy2
![If z = exy2, x = t cost, y = t sint compute Mathematics-II](https://bachelorexam.com/wp-content/uploads/2023/07/image-1211.png)
![If z = exy2, x = t cost, y = t sint compute Mathematics-II](https://bachelorexam.com/wp-content/uploads/2023/07/image-1211.png)
Q5. If cos α, cos β, and cos γ are the direction cosines of a straight line then prove that sin2 α + sin2 β + sin2 γ = 2.
Sol. Since cosα, cosβ, cosγ are the direction cosines of the given line, therefore
![If cos α, cos β, and cos γ are the direction cosines of a straight line then prove](https://bachelorexam.com/wp-content/uploads/2023/07/image-1212.png)
![If cos α, cos β, and cos γ are the direction cosines of a straight line then prove](https://bachelorexam.com/wp-content/uploads/2023/07/image-1212.png)
Section B: Mathematics-II Short Question Solutions
Q6. Show that Dual of a complemented lattice is complemented.
Sol. Let(L, R) be a complemented lattice with 0 and 1 as least and greatest elements. Let (L, R(bar)) be the dual of(L, R) Then 1 and 0 are least and greatest elements of (L, R(bar)).
Let a 𝛜 L be any element.
![Show that Dual of a complemented lattice is complemented.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1213.png)
![Show that Dual of a complemented lattice is complemented.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1213.png)
![Show that Dual of a complemented lattice is complemented.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1214.png)
![Show that Dual of a complemented lattice is complemented.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1214.png)
Q7. Find the equations of the straight line drawn through the origin which will intersect both the lines.
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1215.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1215.png)
Sol. We need to find the equation of line which intersects the lines
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1216.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1216.png)
and passes through (0, 0, 0).
Now,
Equation of line passing through two points (x1,y1, z1) and (x2, y2, z2),
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1217.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1217.png)
We use (x1,y1, z1) = (0, 0, 0)
Now, if point (x2, y2, z2) lies on (L1)
then
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1218.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1218.png)
⇒ x = p + 1, y = 4p – 3, z = 3p + 5
So, (x2, y2, z2) = (p + 1, 4p – 3, 3p + 5)
Thus, the equation of lines will be
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1219.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1219.png)
Now, if the point (x2, y2, z2) lies on (L2)
then,
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1220.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1220.png)
⇒ x = 2q + 4, y = 3q – 3, z = 4q + 14
So, (x2, y2, z2) = (2q + 4, 3q – 3, 4q + 14)
Then the equation of line will be
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1221.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1221.png)
Since, the eqs. (iii) and (iv) represents the same line, so the direction ratios of lines are proportional.
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1222-1024x330.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1222-1024x330.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1223.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1223.png)
Comparing eqs. (v) and (vi),
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1224.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1224.png)
8qk + 16k – 4 = 3qk – 3k + 3
5qk + 19k = 7
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1225.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1225.png)
Comparing eqs. (vi) and (vii),
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1226.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1226.png)
9qk – 9k + 9= 16qk + 56k – 20
7qk + 65 k = 29
k(7q + 65) = 29
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1227.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1227.png)
From eqs. (vii) and (ix),
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1228.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1228.png)
7 (7q + 65) = (5q + 19) 29
49q + 455 = 145 q + 551
96 q = -96
⇒ q = -1
Now, putting the value of q in eq. (iv),
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1229.png)
![Find the equations of the straight line drawn through the origin which will intersect both the lines.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1229.png)
Q8. Show that f(x, y, z) =(x + y + z) – 3(x + y + 2)- 24xyz + a3 has maxima at (-1, – 1, -1).
Sol. f(x, y, z) = (x + y + z)3 – 3(x + y + z)- 24 xyz + a3
Differentiation offw.rtx,y and z,
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1230.png)
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1230.png)
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1231.png)
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1231.png)
3(x + y + z)2 – 3- 24yz = 0
3(x + y + z)2 = 3 + 24yz
(x + y + z)=1 + 8yz …(iv)
and 3(x + y + z)2 – 3 – 24xz = 0
3(x + y + z) = 3 + 24xz
(x + y + z)2 = 1 + 8xz …(v)
and 3(x + y +z)2 – 3 – 24xy = 0
3(x + y + z)2 = 3 + 24xy
(x + y + z) = 3 + 24xy
(x + y + z) = 1 + 8xy …(vi)
From eqs. (iv), (v) and (vi)
1 + 8yz = 1 + 8xz = 1 + 8xy
⇒ x = y = z putting in eq. (iv)
(x + x + x) = 1 + 8xx
(3x)2 = 1 + 8x2
9x2 = 1 + 8x² ⇒ x2 = 1
⇒ x = ± 1
⇒ (x, y, z) = ± (1, 1, 1)
Again,
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1232.png)
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1232.png)
At point (-1, -1, -1),
A = – 18, B = – 18, C = – 18
F = 6, G = 6, H = 6
(1) A = – 18 < 0
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1233.png)
![Show that f(x, y, z) =(x + y + z) - 3(x + y + 2)- 24xyz + a3 has maxima at (-1, - 1, -1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1233.png)
Hence f is maximum at (- 1, – 1,- 1).
Section C: Mathematics-II Detailed Question Solutions
Q9. Let the function f : R → R and g : R → R be defined by f(x) = 2x, g(x) = x2 + 2 ∀x ∈R.
(a) Check the function f and g for being.
(i) One-to-One (ii) Onto
Sol. (1) Given that f(x) = 2x
Checkf is one-one: Let x1, x2 ∈R
Now, f(x) = f(x)
⇒ 2x1 = 2x2
⇒ x1 = x2
Therefore, f is one-one
Check g is one-one: Given that g (x) = x2 + 2
Let x1, x2 ∈R
Now, g(x1) = g(x2)
⇒ x12 +2 = x22 + 2
⇒ x21 = x22
⇒ x21 – x22 = 0
⇒ (x1 – x2) (x1 + x2) = 0
⇒ x1 = ± x2
Therefore, g(x1) = g(x2) does not implies that x1 = x2
Therefore, g is not one- one.
Eg. g(-1) = 1 + 2 = 3
g(1) = 1 + 2 = 3
g(- 1) = g(1)
but – 1 ≠ 1.
(ii) Check f is onto:
Let C ∈ R, f(x) = 2x
⇒ f(x) = C
⇒ 2x= C
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1234.png)
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1234.png)
which implies that C is the image of C/2.
∴ f is onto.
Check g is onto: g(x) = x2 + 2
Let g(x) = C, such that C ∈ R
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1235.png)
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1235.png)
Note that Cis a real number, it can be negative also.
Putting C = – 2
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1236.png)
![Check the function f and g for being.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1236.png)
which is not possible as root of negative number is not real.
Hence x is not real.
So, y is not onto.
(b) Find the formula defining the function fog and gof and obtain the values of (fog) (2) and (gof) (1).
Sol. Composition of function:
Let f: A → B and y : B → C be two real valued functions. Then the composition of f and g denoted by gof, such that gof: A → C is defined by
(gof) (x) =g (f(x))
This is also known as function of a function or resultant of a function.
Similarly, (fog) (x) =f(g(x))
Since we have f(x) = 2x and g(x) = x2 + 2
Now, fog(x) =f(g(x)) = f(x2 + 2) = 2(x2 + 2)
fog (x) = 2(x2 + 2)
fog (2) = 2(22 + 2) = 2(4 + 2) = 12
and gof (x) = g(f (x)) = g(2x) = (2x)2 + 2 = 4x2 + 2
gof (x) = 4x2 +2
gof (1) = 4(1)2 + 2 = 4 + 2 = 6.
Q10. (a) If (L, < ) is a lattice and a, b, c and d ∈ L then.
(i) a ≤ b, c ≤ d ⇒ a ∧ c ≤ b ∧ d
(ii) a ∧ (b v c) ≥ (a ∧ b) v (a ∧ c)
Sol. (i) Since a ∧ c ≤ a and a ∧ c ≤ c,
therefore, again by transitivity
a ∧ c ≤ b and a ∧ c ≤ d (∵ a ≤ b, and c ≤ d).
⇒ a ∧ c is a lower bound of b and d.
But since b ∧ d is the g.l.b of b and d,
therefore, we have a ∧ c ≤ b ∧ d.
(ii) We know that
![If (L, < ) is a lattice and a, b, c and d ∈ L then.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1237.png)
![If (L, < ) is a lattice and a, b, c and d ∈ L then.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1237.png)
because a ∧ (b v c) is the greatest lower bound of {a, b v c}
![If (L, < ) is a lattice and a, b, c and d ∈ L then.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1238.png)
![If (L, < ) is a lattice and a, b, c and d ∈ L then.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1238.png)
(b) Show that dual of a lattice is a lattice.
Sol. Let (L, R) be a given lattice and let (L, R(bar)) be its dual, where R(bar) is defined as x R(bar)y if yRx. Then, it can be shown easily that (L, R(bar)) is a poset.
Let x v y = sup {x, y} in (L, R). Then we have x R(x v y) and y R (x v y)
(x v y) R(bar)x and (x v y) R(bar)y
= x v y is a lower bound of {x, y} in (L, R(bar))
Now, we will show that x v y is the greatest lower bound of {x, y} in (L, R(bar)).
Let z be any lower bound of (x, y) in (L, R(bar)), then z R(bar)x and z R(bar)y.
⇒ x R z and y R z
⇒ z is an upper bound of {x, y} in (L, R)
⇒ (x v y) Rz as x v y = sup {x, y} in (L, R)
⇒ z R(x v y)
⇒ x v y is the greatest lower bound of (x, y) in (L, R(bar)).
Similarly, it can be shown that x ∧ y is the least upper bound in (L, R(bar)).
Therefore (L, R(bar)) is a lattice.
11. (a) Show that f(x, y, z – 2x) = 0, satisfies under suitable conditions, the equation
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1239.png)
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1239.png)
What are these conditions.
Sol. Let u = xy, V = 2 – 2x, then,
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1240-1024x473.png)
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1240-1024x473.png)
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1241.png)
![Show that f(x, y, z - 2x) = 0, satisfies under suitable conditions, the equation](https://bachelorexam.com/wp-content/uploads/2023/07/image-1241.png)
12. (a) Find the equations of the plane parallel to the plane 2x – 3y – 5z + 1 = 0 and distant 5 units from the point (- 1, 3, 1).
Sol. A plane parallel to the plane 2x – 3y – 5z + 1 =0
can be written as
2x – 3y – 5z + 𝛌 = 0
where 𝛌 is a scalar quantity.
Now distance of plane (in eq. (i)) from the point (- 1,3, 1)
![Find the equations of the plane parallel to the plane 2x - 3y - 5z + 1 = 0 and distant 5 units from the point (- 1, 3, 1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1242.png)
![Find the equations of the plane parallel to the plane 2x - 3y - 5z + 1 = 0 and distant 5 units from the point (- 1, 3, 1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1242.png)
Substituting the values of 𝛌 in eq. (i),
![Find the equations of the plane parallel to the plane 2x - 3y - 5z + 1 = 0 and distant 5 units from the point (- 1, 3, 1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1243.png)
![Find the equations of the plane parallel to the plane 2x - 3y - 5z + 1 = 0 and distant 5 units from the point (- 1, 3, 1).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1243.png)
Hence, these are the equations of required planes.
(b) Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x – 6y + 1 = 0 at (1, 2, -2) and passes through the point (1, – 1, 0).
Sol. x2 + y2 + z2 + 2x – 6y + 1 = 0
Equation of tangent plane to the sphere at (1, 2, – 2)
S1 : xx1 + yy1 + zz1 + u(x + x1) + w(y + y1) + v(z2 + z1) + d = 0
S : x2 + y2 + z2 + 2ux + 2wy + 2vz + d = 0
2u = 2, 2w = -6, 2v = 0
u = 1, w = -3, v = 0
S1= 1x + 2y – 2z + 1 (x + 1) + (-3) (y + 2) + 0 + 1 = 0
S1 = 2x + 2y – 2z +1 – 6 – 3y + 1 = 0
= 2x – y – 2z – 4 = 0
Equation of sphere = S + 𝜋 S1 = 0
= (x2 + y2 + z2 + 2x – 6y + 1) + 𝜋 (2x – y – 2z + 4) = 0
Passing through (1, -1),
⇒ (1 +1+0 + 2 +6 + 1) + 𝜋(2 +1-0+4) = 0
⇒ 11 + 7𝜋 = 0
![Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x - 6y + 1 = 0 at (1, 2, -2) and passes through the point (1, - 1, 0).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1244.png)
![Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x - 6y + 1 = 0 at (1, 2, -2) and passes through the point (1, - 1, 0).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1244.png)
Equation of sphere
![Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x - 6y + 1 = 0 at (1, 2, -2) and passes through the point (1, - 1, 0).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1245.png)
![Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x - 6y + 1 = 0 at (1, 2, -2) and passes through the point (1, - 1, 0).](https://bachelorexam.com/wp-content/uploads/2023/07/image-1245.png)
13. (b) Evaluate the following integrals by first converting to Polar coordinates.
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1246.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1246.png)
Sol. The given double integral
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1247.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1247.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1248.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1248.png)
From the limits of integration it is obvious that the region of integration R is bounded by
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1249.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1249.png)
i.e. the region of integration is the area AOBCA of the circle x2 + y2 – 1 = 0 bounded by the lines x = -1 and x = 1.
Putting x = r cos 𝛉,y = r sin 𝛉 the corresponding pol¡r equation of the circle is r2 (cos2𝛉 + sin2𝛉) = 1.
Now, r2 = 1 ⇒ r = 1
Thus, r varies from r = 0 to r = 1 and 𝛉 varies from 𝛉 = 0° to 𝛉 = 𝜋. Also the polar equivalent of dx dy is rd𝛉dr.
Hence, the transformation to polar coordinates,
we have
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1250.png)
![Evaluate the following integrals by first converting to Polar coordinates.](https://bachelorexam.com/wp-content/uploads/2023/07/image-1250.png)