Linear Algebra Questions and Answers – Surface Integrals

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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

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
\(\displaystyle\int_0^a \int_0^{\sqrt{a^2-x^2}}xydxdy \)
\( = \int_0^a \frac{yx^2}{2} dy \) (from 0 to \(\sqrt{a^2-x^2}) \)
\( = \frac{1}{2} \int_0^a y(a^2-y^2) dy= \frac{a^4}{8}. \)
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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

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) \(\frac{1}{12} \)
View Answer

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) \( \frac{\pi}{8} \)
View Answer

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) \(4 \frac{a^2}{6} \)
View Answer

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))
\(= \int_0^π \frac{a^2}{2} (1+cosθ)^2 dθ \)
\(= 4 \frac{a^2}{3}. \)
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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

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.
\(\displaystyle\int_0^a \int_0^\sqrt{a^2-x^2} y\sqrt{x^2-y^2} dxdy \)
a) \( \frac{a^4}{2} \)
b) \( \frac{a^4}{3} \)
c) \( \frac{a^4}{4} \)
d) \( \frac{a^4}{5} \)
View Answer

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
View Answer

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) \( \frac{1}{6} \)
View Answer

Answer: b
Explanation: x: 0 to 1
\(y: x^2 \, to \, x^{1/2} \)
\(\int_0^1 \int_{x^2}^{\sqrt{x}}dydx \)
\(= \int_0^1 \sqrt{x}-x^2 dx \)
\(= \frac{2}{3} – \frac{1}{3} \)
\(= \frac{1}{3}. \)
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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

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 = \( 2* \frac{3πa^2}{4} \)
\(\frac{3πa^2}{2} \).

Sanfoundry Global Education & Learning Series – Linear Algebra.

To practice all areas of Linear Algebra, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn