# 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) $$\frac{a^2}{4}$$

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) $$\frac{a^2}{6}$$

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) $$\frac{1}{12}$$

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) $$\frac{\pi}{8}$$

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) $$4 \frac{a^2}{6}$$

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) $$\frac{\pi}{8}$$

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) $$\frac{a^4}{5}$$

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

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) $$\frac{1}{6}$$

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) $$\frac{3πa^2}{8}$$

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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