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
View Answer
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
View Answer
Explanation: Add constant automatically
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
View Answer
Explanation: Add constant automatically
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) 119⁄22 km
b) 119⁄15 km
c) 129⁄12 km
d) 119⁄12 km
View Answer
Explanation: Add constant automatically
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) 1⁄67
b) 1⁄24
c) –1⁄6
d) –1⁄12
View Answer
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) a4⁄4
b) a4⁄3
c) a5⁄3
d) a2⁄3
View Answer
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) 4⁄3
b) 5⁄3
c) 2⁄3
d) 1
View Answer
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 \(\int\int_0^y \frac{2xy^5}{\sqrt{1+x^2 y^2-y^4}} dxdy\), where y varies from 0 to 1.
a) 11⁄12
b) 14⁄6
c) 11⁄6
d) 11⁄7
View Answer
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) –2⁄3 a4
b) –4⁄3 a4
c) –4⁄3 a5
d) –2⁄3 a5
View Answer
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
View Answer
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) 16⁄946
b) 8⁄945
c) 16⁄45
d) 16⁄945
View Answer
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) 3⁄15
b) 2⁄15
c) 2⁄30
d) 1⁄15
View Answer
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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