This set of Linear Algebra Multiple Choice Questions & Answers (MCQs) focuses on “Volume Integrals”.
1. Find the value of \(\int_0^1 \int_0^2 \int_1^2 xy^2 z^3 dxdydz.\)
a) 2
b) 3
c) 4
d) 5
View Answer
Explanation: \(\int_0^1 \int_0^2 \int_1^2 xy^2 z^3 dxdydz = (\int_0^1 xdx)(\int_0^2 y^2 dy)(\int_1^2 z^3 dz) \)
\(= \frac{1}{2}*\frac{8}{3}*(\frac{16}{4}-\frac{1}{4}) \)
\(= 5. \)
2. Evaluate ∫∫∫z2 dxdydz taken over the volume bounded by the surfaces x2+y2=a2, x2+y2=z and z=0.
a) \(\pi \frac{a^8}{12} \)
b) \(\pi \frac{a^8}{4} \)
c) \(\pi \frac{a^4}{4} \)
d) \(\pi \frac{a^8}{8} \)
View Answer
Explanation: z: 0 to \(x^2+y^2 \)
\( y: \pi \frac{a^8}{12} \) to \( \sqrt{a^2-x^2} \)
x: -a to a
\(\displaystyle\int\int\int z^2 dxdydz = \int_{-a}^a \int_{-\sqrt{(a^2-x^2 )}}^{\sqrt{a^2-x^2}}\int_0^{x^2+y^2}z^2 dxdydz \)
\(\displaystyle = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\frac{z^3}{3} dxdy \) from 0 to \( x^2+y^2. \)
\(\displaystyle = \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{(x^2+y^2 )^3}{3} dxdy\)
\( = \frac{1}{3} \int_0^π \int_{-a}^a r^7 drdθ \)
\( = π \frac{a^8}{12}. \)
3. Find the volume of the cylinder bounded by x2+y2 = 4, y+z = 4 and z=0.
a) \(16π-\frac{32}{3} \)
b) \(32π-\frac{32}{3} \)
c) \(16-32 \frac{π}{3} \)
d) \(32-32 \frac{π}{3} \)
View Answer
Explanation: \(z: (4-y)\)
\(x: -\sqrt{4-y^2} \) to \( \sqrt{4-y^2} \)
y: -2 to 2
Volume = \(\displaystyle\int_{-2}^2 \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} (4-y)dxdy \)
\(= 2\int_{-2}^2 (4y-y^2 )dxdy \) from 0 to \( \sqrt{4-y^2} \)
\(= 4\big\{4(0 + π) – 4 + \frac{4}{3}\big\} \)
\(= 16π-\frac{32}{3}. \)
4. Find the volume of sphere by triple integration.
a) \(8a^3 \frac{π}{3} \)
b) \(4a^3 \frac{π}{3} \)
c) \(2a^3 \frac{π}{3} \)
d) \(a^3 \frac{π}{3} \)
View Answer
Explanation: The sphere is given by x2+y2+z2=a2.
θ : 0 to 2 π
φ : 0 to π
r : 0 to a
Volume = \( \int_0^a \int_0^π \int_0^{2π} r^2 -sinθdθdrdϕ \)
\(= 4(\int_0^a r^2 dr)(\int_0^π sinθdθ)(\int_0^{π/2}dϕ) \)
\(= 4a^3 \frac{π}{3}.\)
5. Using polar coordinates, find the volume of the cylinder with radius a and height h.
a) \(πa^2h \)
b) \(\frac{πa^2h}{3} \)
c) \(2 \frac{πa^2h}{3} \)
d) \( 4 \frac{πa^2h}{3} \)
View Answer
Explanation: r: 0 to a
θ : 0 to 2 π
z: 0 to h
Therefore, volume = \(\int_0^a \int_0^{2π}\int_0^h rdθdrdz \)
\(= (\int_0^a rdr)(\int_0^{2π}dθ)(\int_0^h dz) \)
\(= \frac{a^2}{2} * 2π * h \)
\(= πa^2h. \)
6. In multiple integrals, if the limits depends on variable, then the order of integration can be anything.
a) True
b) False
View Answer
Explanation: In multiple integration, if the limits depend on variable then the order of integration can’t be anything. First the dependent integration should be done and then the independent integration.
7. To find volume _________________ can be used.
a) single integration
b) double integration
c) triple integration
d) double & triple integration
View Answer
Explanation: To find volume, triple integration should be used and proper limits should be given for each variable.
But volume integration can also be done using double integration by using 1D equation of the 3D object as the function.
8. Evaluate \(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z}(x+y+z)dxdydz.\)
a) 0
b) 1
c) 2
d) 3
View Answer
Explanation: \(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z}(x+y+z)dxdydz=
\int_{-1}^1 \int_0^z (xy+\frac{y^2}{2}+zy)dxdz \) from x-z to x+z
\(= \int_{-1}^1\int_0^z(3x^2+z^2+2xz)dxdz= \int_{-1}^1(x^3+z^2 x+x^2 z)dz \) from 0 to z
\(= \int_{-1}^1(z^3+z^3+z^3)dz= \frac{z^4}{4} \) from -1 to 1
= 0.
9. Evaluate \(\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}xyzdxdydz.\)
a) \(\frac{1}{6} \)
b) \(\frac{1}{12} \)
c) \(\frac{1}{24} \)
d) \(\frac{1}{48} \)
View Answer
Explanation: \(\int_0^1 \int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}xyzdxdydz=
\int_0^1 \int_0^{\sqrt{1-x^2}} \frac{xyz^2}{2} dxdy \) from 0 to \( \sqrt{1-x^2-y^2} \)
\(= \frac{1}{2} \int_0^1 \frac{xy^2}{2} – \frac{x}{2} – \frac{xy^4}{4} dx \) from 0 to \(\sqrt{1-x^2} \)
\(= \frac{1}{2} \int_0^1 \frac{(x-x^3)}{2}-\frac{(x^3-x^5)}{2}-\frac{(x+x^5-2x^3)}{4} dx\)
\(= \frac{1}{4}\big\{\frac{1}{6} – \frac{1}{2} + \frac{1}{2}\big\} \)
\(= \frac{1}{24}. \)
10. What is the volume of a cube with side a?
a) \(\frac{a^3}{8} \)
b) \(a^2\)
c) \(a^3\)
d) \(\frac{a2}{4} \)
View Answer
Explanation: \(\int_0^a \int_0^a \int_0^a dxdydz = (\int_0^a dx)(\int_0^a dy)(\int_0^a dz)= a^3.\)
Sanfoundry Global Education & Learning Series – Linear Algebra.
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