Engineering Mathematics Multiple Choice Questions on “Application of Double Integrals”.
1. Distance travelled by any object is _____________
a) Double integral of its acceleration
b) Double integral of its velocity
c) Double integral of its Force
d) Double integral of its Momentum
Answer: b
Explanation: We know that,
x(t) = (intint a(t) ,dtdt).
2. Find the distance travelled by a car moving with acceleration given by a(t)=t2 + t, if it moves from t = 0 sec to t = 10 sec, if velocity of a car at t = 0sec is 40 km/hr.
a) 743.3km
b) 883.3km
c) 788.3km
d) 783.3km
Answer: d
Explanation: Add constant automatically
We know that,
a=(intint(t^2+t)dtdt=intleft [(frac{t^3}{3}+frac{t^2}{2})+cright ]dt)
=(int_0^{10}left [(frac{t^3}{3}+frac{t^2}{2})+40right ]dt=frac{1000}{3}+50+400)=783.3km
3. Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.
a) 10.19 km
b) 19.13 km
c) 15.13 km
d) 13.13 km
Answer: d
Explanation: Add constant automatically
We know that,
a=(intint Sin(t)dtdt=int[-Cos(t)+c]dt=int_0^{π/2}[-Cos(t)+9]dt)
(=[-Sin(t)+9t]=-1+frac{9π}{2})=13.13 km
4. Find the distance travelled by a car moving with acceleration given by a(t)=t2 – t, if it moves from t = 0 sec to t = 1 sec, if velocity of a car at t = 0sec is 10 km/hr.
a) 119⁄22 km
b) 119⁄15 km
c) 129⁄12 km
d) 119⁄12 km
Answer: b
Explanation: Add constant automatically
We know that,
a=(intint[t^2-t] dtdt=int [frac{t^3}{3}-frac{t^2}{2}]dt=int_0^1left [frac{t^3}{3}-frac{t^2}{2}+10 right ] dt=left [frac{t^4}{12}-frac{t^3}{6}+10tright ])
=(frac{1}{12}-frac{1}{6}+10=frac{1}{6} [frac{1}{2}-1]+10 km=-frac{1}{12}+10=frac{119}{12})km
5. Find the value of ∫∫ xydxdy over the area bpunded by parabola y=x2 and x = -y2, is?
a) 1⁄67
b) 1⁄24
c) –1⁄6
d) –1⁄12
Answer: b
Explanation:
(int_0^1 int_{-√y}^{-y^2} y.x dxdy=frac{1}{2} int_0^1 y[y^4-y]dy=frac{1}{2} [frac{1}{6}-frac{1}{3}]=-frac{1}{12})
6. Find the value of (intint ,xydxdy) over the area b punded by parabola x = 2a and x2 = 4ay, is?
a) a4⁄4
b) a4⁄3
c) a5⁄3
d) a2⁄3
Answer: b
Explanation:
(int_0^2a int_0^{x^2/4a} x.y dxdy=int_0^{2a} xdx int_0^{x^2/4a} ydy=int_0^{2a}frac{xx^4}{32a^2}dx)
=(frac{1}{32a^2} int_0^{2a}x^5 dx=frac{a^4}{3})
7. Find the integration of (intint0x (x2 + y2) ,dxdy), where x varies from 0 to 1.
a) 4⁄3
b) 5⁄3
c) 2⁄3
d) 1
Answer: c
Explanation: Add constant automatically
Given, f(x)=(int_0^1 int_0^x (x^2+y^2) dydx = int_0^1 (x^3+frac{x^3}{3})dx=1+frac{1}{3}=frac{2}{3})
8. Evaluate the value of (intint_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy), where y varies from 0 to 1.
a) 11⁄12
b) 14⁄6
c) 11⁄6
d) 11⁄7
Answer: c
Explanation:
Given, f(x)=(int_0^{2a} int_0^y frac{2xy^5}{sqrt{1+x^2 y^2-y^4}} dxdy)
=(int_0^1 int_0^y frac{1}{y} frac{2xy^5}{sqrt{(frac{1-y^4}{y^2})+x^2)}} dxdy=int_0^1 2y^4 |(frac{1-y^4}{y^2})+x^2]|_0^y dy)
=(2int_0^1 [y^3-sqrt{1-y^4}y^3]dy=2left [frac{y^4}{4}-frac{2}{3} (1-y^4)^{frac{3}{2}}right ] _1^0=2[frac{1}{4}+frac{2}{3}]=frac{11}{6})
9. Evaluate ∫∫[x2 + y2 – a2 ]dxdy where, x and y varies from –a to a.
a) –2⁄3 a4
b) –4⁄3 a4
c) –4⁄3 a5
d) –2⁄3 a5
Answer: b
Explanation:
Equation of circle is given by x2 + b2 = a2
(int_{-a}^a int_{-a}^a [x^2+y^2-a^2 ]dxdy)
=(int_{-a}^a [frac{x^3}{3}+y^2 x-a^2 x]_a^{-a}dy=int_{-a}^a [frac{a^3}{3}+ay^2-a^2 a+frac{a^3}{3}+y^2 a-a^2 a]dy)
(int_{-a}^a frac{2a^3}{3}+2ay^2- 2a^3 dy=[frac{2a^3 y}{3}+frac{2ay^3}{3}-2a^3 y] _a^{-a}=frac{4a^4}{3}+frac{4a^4}{3}-4a^4=-frac{4}{3} a^4)
10. Find the area inside a ellipse of minor-radius ‘b’ and major-radius ‘a’.
a) –4⁄3 a2
b) –4⁄3 ab2
c) 4⁄3 ab
d) –4⁄3
Answer: c
Explanation:
Equation of ellipse is given by (frac{x^2}{a^2} +frac{y^2}{b^2}=1)
(int_{-b}^b int_{-a}^a [frac{x^2}{a^2}+frac{y^2}{b^2}-1]dxdy=int_{-b}^b [frac{x^3}{3a^2}+frac{xy^2}{b^2}-x]_a^{-a} dy)
(int_{-b}^b [frac{a}{3}+frac{ay^2}{b^2} -a+frac{a}{3}+frac{ay^2}{b^2}-a] dy)
=(int_{-b}^b [frac{2a}{3}+frac{2ay^2}{b^2}-2a] dy=[frac{2ay}{3}+frac{2ay^3}{3b^2}-2ay] _b^{-b})
=(frac{4ab}{3}+frac{4ab}{3}-4ab=-frac{4}{3} ab)
11. Find the value of (int_0^{1-y} xysqrt{1-x-y} ,dxdy) where, y varies from 0 to 1.
a) 16⁄946
b) 8⁄945
c) 16⁄45
d) 16⁄945
Answer: d
Explanation:
Given, f(x)=(int_1^0∫_0^{1-y} xysqrt{1-x-y} ,dxdy)
putting, (t=frac{x}{1-y})=>x=t(1-y)=>dx=(1-y)dt
(int_1^0int_0^1 t(1-y)ysqrt{1-t(1-y)-y} (1-y)dtdy)
=(int_1^0int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy)
=(int_1^0y(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=β(2,frac{7}{2})β(2,frac{3}{2})=frac{16}{945})
12. Find the value of integral (int_0^1int_{x^2}^x xy(x+y)dydx).
a) 3⁄15
b) 2⁄15
c) 2⁄30
d) 1⁄15
Answer: b
Explanation: Given, F(x)=(int_0^1int_{x^2}^x xy(x+y)dydx=int_0^1 int_{x^2}^x(x ^2 y+xy^2)dydx)
=(int_0^1 [frac{x^2 y^2}{2}+frac{xy^3}{3}] _x^{x^2}dx=int_0^1 [frac{x^3}{2}+frac{x^4}{3}-frac{x^4}{2}-frac{x^5}{3} ]dx=frac{1}{2}+frac{1}{3}-frac{1}{2}-frac{1}{5}=frac{2}{15})