This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses 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) \(\frac{a^2}{4} \)
View Answer
Explanation: In the positive quadrant of the circle,
\(y: 0 \rightarrow a \)
\(x: o \rightarrow \sqrt{a^2-x^2} \)
Therefore the integral is
\(\displaystyle\int_0^a \int_0^{\sqrt{a^2-x^2}}xydxdy \)
\( = \int_0^a \frac{yx^2}{2} dy \) (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) \(\frac{a^2}{6} \)
View Answer
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) \(\frac{1}{12} \)
View Answer
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) \( \frac{\pi}{8} \)
View Answer
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) \(4 \frac{a^2}{6} \)
View Answer
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) \( \frac{\pi}{8} \)
View Answer
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.
\(\displaystyle\int_0^a \int_0^\sqrt{a^2-x^2} y\sqrt{x^2-y^2} dxdy \)
a) \( \frac{a^4}{2} \)
b) \( \frac{a^4}{3} \)
c) \( \frac{a^4}{4} \)
d) \( \frac{a^4}{5} \)
View Answer
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
View Answer
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) \( \frac{1}{6} \)
View Answer
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) \(\frac{3πa^2}{8} \)
View Answer
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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