250+ TOP MCQs on Double Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Double Integrals”.

1. Find the value of ∫∫xyex + y dxdy.
a) yey (xex-ex)
b) (yey-ey)(xex-ex)
c) (yey-ey)xex
d) (yey-ey)(xex+ex)
Answer: b
Add constant automatically
Explanation: Given, ∫∫xyex + y dxdy
∫∫xyex ey dxdy= ∫yey dy∫xex dx=(yey-ey)(xex-ex).

2. Find the value of ∫∫ xx2 + y2 dxdy.
a) [ytan(-1) (y)- 12 ln⁡(1+y2)]
b) x [ytan(-1) (y)- 12 ln⁡(1+y2)]
c) y [xtan(-1) (x)- 12 ln⁡(1+x2)]
d) x [ytan(-1) (y)- 12 ln⁡(1+y2)]
Answer: d
Explanation: Add constant automatically
Given, (intint 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 = xint 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)- 12 ln⁡(1+y2)].

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))
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 (intint_0^{sqrt{2ax-x^2}}x ,dxdx).
a) ax22x530
b) ax22x36
c) ax22
d) ax48x36
Answer: b
Explanation: Add constant automatically
Given, f(x)=(intint_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))
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 ∫∫0x x2 + y2 dxdy.
a) x46
b) y
c) 2x33y
d) 1
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 (intint_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}}])
Answer: c
Explanation:
Given, f(x)=(intint_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy)
=(intint_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^2right |_0^y dy)
(=2int [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 (intint_0^{1-y} xysqrt{1-x-y} ,dxdy).
a) 16946
b) 8945
c) 16936
d) 16945
Answer: d
Explanation:
Given, f(x)=(int_0^1 int_0^{1-y} xysqrt{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)ysqrt{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} dyint_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) a22 + 5a – 4ln(a)
b) a22 + 5a – 4ln(a) – 112
c) a22 + 4ln(a) – 112
d) a22 + 5a – 112
Answer: b
Explanation: Add constant automatically
Given,
(f(x) = frac{(2x^3+5x^2-4)}{x^2})

Integrating it we get, F(x) = x22 + 5x – 4ln(x)

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

F(a) – F(1) = a22 + 5a – 4ln(a) – 12 – 5 = a22 + 5a – 4ln(a) – 112.

10. Find the value of (intintfrac{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))
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))

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