Double Integrals - Engineering Mathematics Questions and Answers

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Double Integrals”.

1. Find the value of ∫∫xye^{x + y} dxdy.
a) ye^y (xe^x-e^x)
b) (ye^y-e^y)(xe^x-e^x)
c) (ye^y-e^y)xe^x
d) (ye^y-e^y)(xe^x+e^x)
View Answer

Answer: b
Add constant automatically
Explanation: Given, ∫∫xye^{x + y} dxdy
∫∫xye^x e^y dxdy= ∫ye^y dy∫xe^x dx=(ye^y-e^y)(xe^x-e^x).

2. Find the value of ∫∫ ^x⁄_{x² + y²} dxdy.
a) [ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y²)]
b) x [ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y²)]
c) y [xtan^(-1) (x)- ¹⁄₂ ln⁡(1+x²)]
d) x [ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y²)]
View Answer

Answer: d
Explanation: Add constant automatically
Given, \(\int\int \frac{x}{x^2+y^2} \,dxdy\)
\(\int x \int \frac{1}{x^2+y^2} \,dydx=\int x \frac{1}{x} tan^{-1}⁡(\frac{y}{x}) \,dy=\int tan^{-1}⁡(\frac{y}{x}) \,dy\)
\(\int tan^{-1}⁡(\frac{y}{x})\,dy = x\int tan^{-1}⁡(t)\,dt\)
Putting, x = tan(z),
We get, dz = sec²⁡(z)dz,

x∫ zsec² (z)dz

By integration by parts,

x ∫ zsec² (z)dz=x[ztan(z)-log⁡(sec⁡(z))]= x[ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y²)].

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3. Find the ∫∫x³ y³ sin⁡(x)sin⁡(y) dxdy.
a) (x³ Cos(x) + 3x² Sin(x) + 6xCos(x)-6Sin(x))(y³ Cos(y) + 3[y² Sin(y) – 2[-yCos(y) + Sin(y)]])
b) (-x³ Cos(x) – 3x² Sin(x) – 6xCos(x)-6Sin(x))(-y³ Cos(y) + 3[y² Sin(y) – 2[-yCos(y) + Sin(y)]])
c) (-x³ Cos(x) + 3x² Sin(x) + 6xCos(x)-6Sin(x))(-y³ Cos(y) + 3y² Sin(y) + 6yCos(y) – 6Sin(y))
d) (–x³ Cos(x) + 6xCos(x) – 6Sin(x))(-y³ Cos(y))
View Answer

Answer: c
Explanation: Add constant automatically
∫x³ Sin(x)dx = -x³ Cos(x) + 3∫x² Cos(x)dx
∫x² Cos(x)dx = x² Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x)
=> ∫x³ Sin(x)dx = -x³ Cos(x) + 3[x² Sin(x) – 2[-xCos(x) + Sin(x)]] => ∫x³ Sin(x)dx = -x³ Cos(x) + 3x² Sin(x) + 6xCos(x) – 6Sin(x)
and, ∫y³ Sin(y)dy = -y³ Cos(x) + 3∫y² Cos(y)dy
∫y² Cos(y)dy = y² Sin(y) – 2∫ySin(y)dy
∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y)
=> ∫y³ Sin(y)dy = -y³ Cos(y) + 3[y² Sin(y) – 2[-yCos(y) + Sin(y)]] => ∫y³ Sin(y)dy = -y³ Cos(y) + 3y² Sin(y) + 6yCos(y) – 6Sin(y)
Hence, ∫∫x³ y³ sin⁡(x) sin⁡(y) dxdy = (∫x³ Sin(x)dx)(∫y³ Sin(y)dy) = (-x³ Cos(x) + 3x² Sin(x)+6xCos(x) – 6Sin(x))(-y³ Cos(y) + 3y² Sin(y) + 6yCos(y) – 6Sin(y)).

4. Find the integration of \(\int\int_0^{\sqrt{2ax-x^2}}x \,dxdx\).
a) ^ax²⁄₂ – ^x⁵⁄₃₀
b) ^ax²⁄₂ – ^x³⁄₆
c) ^ax²⁄₂
d) ^ax⁴⁄₈ – ^x³⁄₆
View Answer

Answer: b
Explanation: Add constant automatically
Given, f(x)=\(\int\int_0^{\sqrt{2ax-x^2}}x \,dxdx = \int [\frac{x^2}{2}]_0^{\sqrt{2ax-x^2}} \,dxdx = \int \frac{2ax-x^2}{2} \,dx=\frac{ax^2}{2}-\frac{x^3}{6}\)

5. Find the value of ∫∫xy⁷ Cos(x)Cos(y) dxdy.
a) (7y⁶ Cos(y) + 42y⁵ Sin(y) + 210y⁴ Cos(y) + 840y³ Sin(y) + 2520y² Cos(y) + 5040ySin(y) + 5040Cos(y))(7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x))
b) (y⁷ Sin(y) + 7y⁶ Cos(y) + 42y⁵ Sin(y) + 210y⁴ Cos(y) + 840y³ Sin(y) + 2520y² Cos(y) + 5040ySin(y) + 5040Cos(y))(x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x))
c) (y⁷ Sin(y) + 42y⁵ Sin(y) + 210y⁴ Cos(y) + 840y³ Sin(y) + 2520y² Cos(y) + 5040ySin(y) + 5040Cos(y))(x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x))
d) (y⁷ Sin(y) + 7y⁶ Cos(y) + 42y⁵ Sin(y) + 210y⁴ Cos(y) + 840y³ Sin(y) + 2520y² Cos(y) + 5040ySin(y) + 5040Cos(y))(x⁷ Sin(x) + 7x⁶ Cos(x) + 42x⁵ Sin(x) + 210x⁴ Cos(x) + 840x³ Sin(x) + 2520x² Cos(x) + 5040xSin(x) + 5040Cos(x))
View Answer

Answer: d
Explanation: Add constant automatically
By, f(x)=\(\int uvdx=\sum_{i=0}^n (-1)^i u_i v^{i+1}\),
Let, u = x⁷ and v=Cos(x),
∫x⁷ Cos(x) dx=x⁷ Sin(x)+7x⁶ Cos(x)+42x⁵ Sin(x)+210x⁴ Cos(x)+840x³ Sin(x)+2520x² Cos(x)+5040xSin(x)+5040Cos(x)
Similarly,
∫y⁷ Cos(y) dy=y⁷ Sin(y)+7y⁶ Cos(y)+42y⁵ Sin(y)+210y⁴ Cos(y)+840y³ Sin(y)+2520y² Cos(y)+5040ySin(y)+5040Cos(y)
Now,
∫∫xy⁷ Cos(x)Cos(y) dxdy=∫y⁷ Cos(y) dy∫x⁷ Cos(x) dx=(y⁷ Sin(y)+7y⁶ Cos(y)+42y⁵ Sin(y)+210y⁴ Cos(y)+840y³ Sin(y)+2520y² Cos(y)+5040ySin(y)+5040Cos(y))(x⁷ Sin(x)+7x⁶ Cos(x)+42x⁵ Sin(x)+210x⁴ Cos(x)+840x³ Sin(x)+2520x² Cos(x)+5040xSin(x)+5040Cos(x)).

6. Find the integration of ∫∫₀^x x² + y² dxdy.
a) ^x⁴⁄₆
b) y
c) ^2x³⁄₃y
d) 1
View Answer

Answer: c
Explanation: Add constant automatically
Given, \(f(x)=∫_0^x (x^2+y^2) \,dxdy = ∫ (\frac{x^3}{3}+\frac{x^3}{3})\,dxdy = \frac{2x^3}{3} y\).

7. Find the value of \(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\).
a) \(2[\frac{y^4}{4}- \frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)
b) \(2[\frac{y^4}{4}- (1-y^4)^{\frac{3}{2}}]\)
c) \(2[\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)
d) \(2[\frac{y^3}{3}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}]\)
View Answer

Answer: c
Explanation:
Given, f(x)=\(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\)
=\(\int\int_0^y \frac{1}{y} \frac{2xy^5}{\sqrt{(\frac{1-y^4}{y^2})+x^2}} dxdy=\int 2y^4 \left |(\frac{1-y^4}{y^2})+x^2\right |_0^y dy\)
\(=2\int [y^3-\sqrt{1-y^4}y^3]dy=2[\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{3/2}]\)

8. Find the value of \(\int\int_0^{1-y} xy\sqrt{1-x-y} \,dxdy\).
a) ¹⁶⁄₉₄₆
b) ⁸⁄₉₄₅
c) ¹⁶⁄₉₃₆
d) ¹⁶⁄₉₄₅
View Answer

Answer: d
Explanation:
Given, f(x)=\(\int_0^1 \int_0^{1-y} xy\sqrt{1-x-y} dxdy\)
putting,t=x/(1-y)=>x=t(1-y)=>dx=(1-y)dt
\(\int_0^1 \int_0^1 t(1-y)y\sqrt{1-t(1-y)-y} (1-y)dtdy\)
=\(\int_0^1 \int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy\)
=\(\int_0^1y(1-y)^{5/2} dy \int_0^1 t(1-t)^{1/2} dt\)
=\(\int_0^1 y^{2-1} (1-y)^{7/2-1} dy\int_0^1 t^{2-1} (1-t)^{3/2-1} dt=\beta(2,\frac{7}{2})\beta(2,\frac{3}{2})=\frac{16}{945}\)

9. Find the area inside function ^{(2x³ + 5 x² – 4)}⁄_x² from x = 1 to a.
a) ^a²⁄₂ + 5a – 4ln(a)
b) ^a²⁄₂ + 5a – 4ln(a) – ¹¹⁄₂
c) ^a²⁄₂ + 4ln(a) – ¹¹⁄₂
d) ^a²⁄₂ + 5a – ¹¹⁄₂
View Answer

Answer: b
Explanation: Add constant automatically
Given,
\(f(x) = \frac{(2x^3+5x^2-4)}{x^2}\)

Integrating it we get, F(x) = ^x²⁄₂ + 5x – 4ln(x)

Hence, area under, x = 1 to a, is

F(a) – F(1) = ^a²⁄₂ + 5a – 4ln(a) – ¹⁄₂ – 5 = ^a²⁄₂ + 5a – 4ln(a) – ¹¹⁄₂.

10. Find the value of \(\int\int\frac{1}{16x^2+16x+10} \,dx\).
a) \(\frac{1}{8} (x+\frac{1}{2})Sin^{-1} (x+\frac{1}{2})-\frac{1}{2} ln⁡(1+(x+\frac{1}{2})^2)\)
b) \(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
c) \((x+1/2) cos^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
d) \(\frac{1}{8} (x+\frac{1}{2}) sec^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)
View Answer

Answer: b
Explanation: Add constant automatically
Given,\(\int \frac{1}{16x^2+16x+10} dx=\frac{1}{2}\int \frac{1}{4x^2+4x+5} dx\)
=\(\int \frac{1}{8(x^2+x+5/4+1/4+1/4)} dx\)
=\(\int \frac{1}{8[(x+1/2)^2+1^2]}dx=\frac{1}{8} tan^{-1}⁡(x+1/2)\)
Hence, \(\frac{1}{8} \int tan^{-1}⁡(x+\frac{1}{2})dx\)
Now, Putting, x+1/2 = tan(y),
We get, dx = sec²⁡(y)dy,
=1/8 \(\int ysec^2 (y)dy\)
By integration by parts,
ytan(y)-log⁡(sec⁡(y))=\(\frac{1}{8} (x+\frac{1}{2})tan^{-1} (x+\frac{1}{2})-\frac{1}{16} ln⁡(1+(x+\frac{1}{2})^2)\)

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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