Engineering Mathematics Questions and Answers – Application of Double Integrals

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses 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

Explanation: We know that,
x(t) = $$\int\int a(t) \,dtdt$$.

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

We know that,
a=$$\int\int(t^2+t)dtdt=\int\left [(\frac{t^3}{3}+\frac{t^2}{2})+c\right ]dt$$
=$$\int_0^{10}\left [(\frac{t^3}{3}+\frac{t^2}{2})+40\right ]dt=\frac{1000}{3}+50+400$$=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

We know that,
a=$$\int\int Sin(t)dtdt=\int[-Cos(t)+c]dt=\int_0^{π/2}[-Cos(t)+9]dt$$
$$=[-Sin(t)+9t]=-1+\frac{9π}{2}$$=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) 11922 km
b) 11915 km
c) 12912 km
d) 11912 km

We know that,
a=$$\int\int[t^2-t] dtdt=\int [\frac{t^3}{3}-\frac{t^2}{2}]dt=\int_0^1\left [\frac{t^3}{3}-\frac{t^2}{2}+10 \right ] dt=\left [\frac{t^4}{12}-\frac{t^3}{6}+10t\right ]$$
=$$\frac{1}{12}-\frac{1}{6}+10=\frac{1}{6} [\frac{1}{2}-1]+10 km=-\frac{1}{12}+10=\frac{119}{12}$$km

5. Find the value of ∫∫ xydxdy over the area bpunded by parabola y=x2 and x = -y2, is?
a) 167
b) 124
c) –16
d) –112

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 $$\int\int \,xydxdy$$ over the area b punded by parabola x = 2a and x2 = 4ay, is?
a) a44
b) a43
c) a53
d) a23

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 $$\int\int0x (x2 + y2) \,dxdy$$, where x varies from 0 to 1.
a) 43
b) 53
c) 23
d) 1

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 $$\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy$$, where y varies from 0 to 1.
a) 1112
b) 146
c) 116
d) 117

Explanation:
Given, f(x)=$$\int_0^{2a} \int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy$$
=$$\int_0^1 \int_0^y \frac{1}{y} \frac{2xy^5}{\sqrt{(\frac{1-y^4}{y^2})+x^2)}} dxdy=\int_0^1 2y^4 |(\frac{1-y^4}{y^2})+x^2]|_0^y dy$$
=$$2\int_0^1 [y^3-\sqrt{1-y^4}y^3]dy=2\left [\frac{y^4}{4}-\frac{2}{3} (1-y^4)^{\frac{3}{2}}\right ] _1^0=2[\frac{1}{4}+\frac{2}{3}]=\frac{11}{6}$$

9. Evaluate ∫∫[x2 + y2 – a2 ]dxdy where, x and y varies from –a to a.
a) –23 a4
b) –43 a4
c) –43 a5
d) –23 a5

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) –43 a2
b) –43 ab2
c) 43 ab
d) –43

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 $$\int_0^{1-y} xy\sqrt{1-x-y} \,dxdy$$ where, y varies from 0 to 1.
a) 16946
b) 8945
c) 1645
d) 16945

Explanation:
Given, f(x)=$$\int_1^0∫_0^{1-y} xy\sqrt{1-x-y} \,dxdy$$
putting, $$t=\frac{x}{1-y}$$=>x=t(1-y)=>dx=(1-y)dt
$$\int_1^0\int_0^1 t(1-y)y\sqrt{1-t(1-y)-y} (1-y)dtdy$$
=$$\int_1^0\int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy$$
=$$\int_1^0y(1-y)^{5/2} dy \int_0^1 t(1-t)^{1/2} dt$$
=$$\int_0^1 y^{2-1} (1-y)^{7/2-1} dy \int_0^1 t^{2-1} (1-t)^{3/2-1} dt=β(2,\frac{7}{2})β(2,\frac{3}{2})=\frac{16}{945}$$

12. Find the value of integral $$\int_0^1\int_{x^2}^x xy(x+y)dydx$$.
a) 315
b) 215
c) 230
d) 115

Explanation: Given, F(x)=$$\int_0^1\int_{x^2}^x xy(x+y)dydx=\int_0^1 \int_{x^2}^x(x ^2 y+xy^2)dydx$$
=$$\int_0^1 [\frac{x^2 y^2}{2}+\frac{xy^3}{3}] _x^{x^2}dx=\int_0^1 [\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^4}{2}-\frac{x^5}{3} ]dx=\frac{1}{2}+\frac{1}{3}-\frac{1}{2}-\frac{1}{5}=\frac{2}{15}$$

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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