Linear Algebra Multiple Choice Questions on “Surface Integrals”.
1. Evaluate ∫xy dxdy over the positive quadrant of the circle x2+y2=a2.
a) (frac{a^4}{8} )
b) (frac{a^4}{4} )
c) (frac{a^2}{8} )
d) (frac{a^2}{4} )
Answer: a
Explanation: In the positive quadrant of the circle,
(y: 0 rightarrow a )
(x: o rightarrow sqrt{a^2-x^2} )
Therefore the integral is
(displaystyleint_0^a int_0^{sqrt{a^2-x^2}}xydxdy )
( = int_0^a frac{yx^2}{2} dy ) (from 0 to (sqrt{a^2-x^2}) )
( = frac{1}{2} int_0^a y(a^2-y^2) dy= frac{a^4}{8}. )
2. Evaluate ∫∫xy dxdy over the region bounded by x axis, ordinate x=2a and the curve x2=4ay.
a) (frac{a^4}{3} )
b) (frac{a^4}{6} )
c) (frac{a^3}{3} )
d) (frac{a^2}{6} )
Answer: a
Explanation: Both the curves meet at (2a,a).
Therefore,
(x:0 rightarrow 2a )
(y: 0 rightarrow frac{x^2}{4a} )
(int_0^2 a int_0^{frac{x^2}{4}}a xydxdy )
(= int_0^2 a frac{xy^2}{2} dx ) (from 0 to ( frac{x^2}{4a}) )
( = frac{1}{2} int_0^{2a} frac{x^5}{(16a^2 )} dx )
( = frac{a^4}{3}. )
3. Evaluate ∫∫x2+y2 dxdy in the positive quadrant for which x+y<=1.
a) (frac{1}{2} )
b) (frac{1}{3} )
c) (frac{1}{6} )
d) (frac{1}{12} )
Answer: c
Explanation: In this
x: 0 to 1
y:0 to 1-x
(int_0^1 int_0^{1-x} x^2+y^2 dxdy )
(= int_0^1 x^2 y+ frac{y^3}{3} dx ) (from 0 to 1-x)
(= int_0^1 x^2 (1-x)+ frac{(1-x)^3}{3} dx )
(= frac{1}{6}. )
4. Evaluate (int_0^∞ int_0^{π/2} e^{-r^{2}} rdθdr ).
a) ( pi )
b) ( frac{pi}{2} )
c) ( frac{pi}{4} )
d) ( frac{pi}{8} )
Answer: c
Explanation: The integral is in polar coordinates.
Substitute r2 as t
(int_0^∞ int_0^{π/2} e^{-t} dθ frac{dt}{2} )
(= frac{1}{2} int_0^{π/2}Γ(1)dθ )
(= frac{pi}{4}. )
5. Evaluate ∫∫rsinθdrdθ over the cardiod r = a(1+cosθ) above the initial line.
a) (4 frac{a^2}{3} )
b) ( frac{a^2}{3} )
c) (8 frac{a^2}{3} )
d) (4 frac{a^2}{6} )
Answer: a
Explanation: θ: 0 to π
r: 0 to a(1+cosθ)
(int_0^π int_0^{a(1+cosθ)} rsinθdrdθ )
(= int_0^π frac{r^2}{2} sinθdθ ) (from 0 to a(1+cosθ))
(= int_0^π frac{a^2}{2} (1+cosθ)^2 dθ )
(= 4 frac{a^2}{3}. )
6. Evaluate (int_0^∞ int_0^∞ e^{-(x^2+y^2 )} dxdy ) by changing into polar coordinates.
a) ( pi )
b) ( frac{pi}{2} )
c) ( frac{pi}{4} )
d) ( frac{pi}{8} )
Answer: c
Explanation: ( int_0^∞ int_0^∞ e^{-(x^2+y^2 )} dxdy )
(= int_0^{π/2} int_0^∞ e^{-(r^2 )} drdθ )
Substitute (r^2) as t
(= frac{1}{2} int_0^{π/2} int_0^∞ e^{-t} dtdθ )
( = frac{1}{2} int_0^{π/2}Γ(1)dθ )
( = frac{pi}{4}. )
7. Evaluate the following integral by transforming into polar coordinates.
(displaystyleint_0^a int_0^sqrt{a^2-x^2} ysqrt{x^2-y^2} dxdy )
a) ( frac{a^4}{2} )
b) ( frac{a^4}{3} )
c) ( frac{a^4}{4} )
d) ( frac{a^4}{5} )
Answer: c
Explanation: Subtitute x as rcosθ and y as rsinθ.
Therfore θ : 0 to Π/2
and r : 0 to a
(int_0^a int_0^{π/2} rsinθrrdrdθ )
(= [int_0^a r^3 dr][int_0^{π/2}sinθdθ] )
(= frac{a^4}{4} ).
8. Evaluate (int_0^∞ int_x^∞ frac{e^{-y}}{y} dydx ) by changing the order of integration.
a) 0
b) 1
c) 2
d) 1/2
Answer: b
Explanation: In the question, y: x to infnity
x: 0 to infinity
Now changing the orrder of integration:
y=x
y tends to infinity
y: 0 to infinity
x: 0 to y
(int_0^∞ int_0^y frac{e^{-y}}{y} dydx )
( = int_0^∞ frac{e^{-y}}{y} ydy )
= -(0-1)
= 1.
9. Calculate the area enclosed by parabolas x2 = y and y2 = x.
a) ( frac{1}{2} )
b) ( frac{1}{3} )
c) ( frac{1}{4} )
d) ( frac{1}{6} )
Answer: b
Explanation: x: 0 to 1
(y: x^2 , to , x^{1/2} )
(int_0^1 int_{x^2}^{sqrt{x}}dydx )
(= int_0^1 sqrt{x}-x^2 dx )
(= frac{2}{3} – frac{1}{3} )
(= frac{1}{3}. )
10. What is the area of a cardiod y = a(1+cosθ).
a) (frac{3πa^2}{2} )
b) (3πa^2 )
c) (frac{3πa^2}{4} )
d) (frac{3πa^2}{8} )
Answer: a
Explanation: θ : 0 to π
r : 0 to a(1+cosθ)
(frac{Area}{2} = int_0^π int_0^{a(1+cosθ)}rdrdθ )
(= frac{3πa^2}{4} )
Total area = ( 2* frac{3πa^2}{4} )
(frac{3πa^2}{2} ).
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