PYQs on Multivariable Calculus - 1 | Engineering Mathematics

In examinations, questions from Multivariable calculus are frequently asked in the form of evaluating double and triple integrals, changing the order of integration, as well as problems based on change of variables.

Short Question on Multivariable Calculus - 1

Question 1: Evaluate I = \int_{2}^{3} \int_{-1}^{4} \int_{0}^{1} \left( 4x^2 y - z^3 \right) \, dz \, dy \, dx

Question 2: Evaluate\iiint_{0 \leq x,y,z \leq 1} (x^2+y^2+z^2)\, dx \ dy \ dz.

Question 3: Find the volume of the solid obtained by rotating the region bounded by y =\sqrt{x}, y = 0, and x = 4 about the x-axis.

Question 4: Find the Jacobian of the transformation x = rcos⁡θ,  y = rsin⁡θ.

Question 5: Find the volume bounded by the paraboloid z = x²+y² and the plane z = 9.

Question 6: Change the order of integration:\int_0^2 \int_0^{\sqrt{4-y^2}} (x^2+y^2)\, dx \ dy

Question 7: Find the volume of the sphere x²+y²+z² ≤ a² using spherical coordinates.

Question 8: Find the volume of the solid obtained by revolving the region bounded by y = x and y = x² about the x-axis, over [0, 1].

Question 9: Find the average value of f(x) = x³ on [0, 2].

Question 10: A particle moves with velocity v(t) = 10 + 2t m/s for 0 ≤ t ≤ 50. Find displacement.

Long Questions on Multivariate Calculus - 1

Question 11: Find the volume of the solid bounded by the cylinder x² + y² = 4x² + y², the planes z = 0 and z = 5, using triple integration in cylindrical coordinates.

Question 12: Find the volume of the solid bounded by the cylinder x²+ y² = 1 and planes z = 0, z = h.

Check if you were right - full answer with solution below.

Solution 1:

I = \int_{2}^{3} \int_{-1}^{4} \int_{0}^{1} \left( 4x^2 y - z^3 \right) \, dz \, dy \, dx

Integrate w.r.t z:

\int_{0}^{1} \left( 4x^2 y - z^3 \right) \, dz = 4x^2 y \cdot \Big[ z \Big]_0^1 - \Big[ \frac{z^4}{4} \Big]_0^1= 4x^2 y - \frac{1}{4}

Integrate w.r.t y:

\int_{-1}^{4} \left( 4x^2 y - \frac{1}{4} \right) \, dy

= \left[ 2 x^2 y^2 - \frac{1}{4}y \right]_{-1}^{4}

= 2x²( 16 - 1) - 5/4

= 30x² - 5/4

Integrate w.r.t x:

\int_{2}^{3} \left( 30x^2 - \frac{5}{4} \right) \, dx

= \left[ 10x^3 - \frac{5}{4}x \right]_{2}^{3}

= (270 - 15/4) - 80 - 10/4) = 190 - 5/4 = 755/4

Solution 2:

\iiint_{0 \leq x,y,z \leq 1} (x^2+y^2+z^2)\, dxdydz

Separate the integral:

\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (x^2 + y^2 + z^2) \, dz \, dy \, dx= \int_{0}^{1} \int_{0}^{1} \left( \int_{0}^{1} x^2 \, dz + \int_{0}^{1} y^2 \, dz + \int_{0}^{1} z^2 \, dz \right) \, dy \, dx

• \int_0^1 x^2 dz = x²
• \int_0^1 y^2 dz = y²
• \int_0^1 z^2 dz = 1/3

So integral reduces to:

\int_{0}^{1} \int_{0}^{1} \left( x^2 + y^2 + \frac{1}{3} \right) \, dy \, dx

= \int_{0}^{1} \left( x^2 + \frac{1}{3} + \int_{0}^{1} y^2 \, dy \right) \, dx

= \int_{0}^{1} \left( x^2 + \frac{1}{3} + \frac{1}{3} \right) \, dx

= \int_{0}^{1} \left( x^2 + \frac{2}{3} \right) \, dx

\int_{0}^{1} x^2 \, dx + \int_{0}^{1} \frac{2}{3} \, dx= \frac{1}{3} + \frac{2}{3} = 1

Solution 3:

Washer method formula:

V = \pi \int_{0}^{4} \left[ x^2 - 0^2 \right] \, dx = \pi \int_{0}^{4} x^2 \, dx

Solve integral:

V = \pi \int_{0}^{4} x \, dx
= pi \left[ \frac{x^2}{2} \right]_0^4
= 16/2 = 8

V = π⋅8 = 8π
V = 8π

Solution 4:

Jacobian formula:

J = \frac{\partial (r, \theta)}{\partial (x, y)}=\begin{vmatrix}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\\frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y}\end{vmatrix}

Compute partial derivatives:

\frac{\partial r}{\partial x} = \cos \theta, \quad\frac{\partial \theta}{\partial x} = -r \sin \theta, \quad\frac{\partial r}{\partial y} = \sin \theta, \quad\frac{\partial \theta}{\partial y} = r \cos \theta

\begin{vmatrix}\cos \theta & \sin \theta \\- r \sin \theta & r \cos \theta\end{vmatrix} = r( cos²θ+sin2θ) = r

Solution 5:

Convert to polar coordinates:

x = rcos⁡θ,  y = rsin⁡θ,  z = r² → 9

• Limits: 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π , z = r² → 9

Volume formula:

V = \int_{0}^{2\pi} \int_{0}^{3} \int_{r^2}^{9} dz \, r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{3} (9 - r^2) \, r \, dr \, d\theta

\int_{0}^{3} (9r - r^3) \, dr

= \left[ \tfrac{9}{2}r^2 - \tfrac{1}{4}r^4 \right]_{0}^{3}

= 81/2 - 81/4 = 81/4

V = \int_{0}^{2\pi} \frac{81}{4} \, d\theta = \frac{81}{4} \cdot (2\pi)= \frac{81}{2} \pi.

Solution 6:

Identify region

• Inner limit: 0 ≤ x ≤\sqrt{ 4−y2}
• Outer limit: 0 ≤ y ≤ 2
• This represents quarter-circle x² + y² ≤ 4 in the first quadrant.

Reverse the limits

• For fixed x: y goes from 0 to \sqrt{ 4−y^2}
• For x: range is 0 ≤ x ≤ 2.

I = \int_{0}^{2} \int_{0}^{4 - x^2} \left( x^2 + y^2 \right) \, dy \, dx

Solution 7:

In spherical coordinates

x = ρsin⁡ϕcos⁡θ,  y = ρsin⁡ϕsin⁡θ,  z = ρcos⁡ϕ

Jacobian: dV = ρ²sin⁡ϕ dρ dϕ dθdV = ρ²sinϕdρdϕdθ.

Limits

0 ≤ ρ ≤ a,  0 ≤ ϕ ≤ π,    0≤ θ ≤ 2π

Integral

V = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \rho^{2} \sin\phi \, d\rho \, d\phi \, d\theta

Solve

\int_{0}^{a} \rho^{2} \, d\rho = \frac{a^{3}}{3},
\int_{0}^{\pi} \sin\phi \, d\phi = 2,
\int_{0}^{2\pi} d\theta = 2\pi

V = a³/3 . 2 . 2π = 4/3πa³

Solution 8:

V = \pi \int_{a}^{b} \left( y_{\text{upper}}^2 - y_{\text{lower}}^2 \right) \, dx

V = \pi \int_{0}^{1} \left( x^2 - (x^2)^2 \right) dx

= \pi \int_{0}^{1} \left( x^2 - x^4 \right) dx

Integrate:

1/3 - 1/5 = 2/15

V = π . 2/15 = 2/15 π

Solution 9:

f_{\text{avg}} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx

f_{\text{avg}} = \frac{1}{2-0} \int_{0}^{2} x^3 \, dx

= \frac{1}{2} \left[ \frac{x^4}{4} \right]_{0}^{2}

= 1/2 . 4 = 2

Solution 10:

Displacement:

s = \int_{0}^{50} v(t) \, dt

= \int_{0}^{50} ( 10 + 2t) dt

= \int_{0}^{50} 10 \, dt + \int_{0}^{50} 2t \, dt

= 10 . 50 + 2. 50²/2 = 500 + 2500 = 3000

Solution 11:

Convert cylinder equation:

x² + y² = 4x² + y

0 = 3x²

x = 0

Cylindrical coordinates:

x = rcosθ, y = rsinθ, z = z

• r limits: 0 to some radius R
• θ : 0 → 2π
• z : 0 → 5

Volume integral:

V = \int_{0}^{2\pi} \int_{0}^{R} \int_{0}^{5} r \, dz \, dr \, d\theta

Solve:

Integrate w.r.t z: \int_0^5 dz = 5

V = \int_{0}^{2\pi} \int_{0}^{R} 5r \, dr \, d\theta

Integrate w.r.t r: \int_0^{R} 5rdr = 5R²/2

Integrate w.r.t θ: \int_0^{2\pi} d\theta = 2\pi

V = 5R²/2 . 2π = 5πR²
V = 5πR²

Solution 12:

Use cylindrical coordinates:
x = rcos⁡θ, y = rsin⁡θ, z = z

• Cylinder: x² + y² = r² = 1  ⟹  r ∈ [0,1]
• Angle: θ ∈ [0, 2π]
• Height: z ∈ [0, h]

Volume element: dV = r dr dθ dz

Set up the integral

\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{h} r \, dz \, dr \, d\theta

Integrate with respect to z:

\int_{0}^{h} dz = h
V = \int_{0}^{2\pi} \int_{0}^{1} h r \, dr \, d\theta = h \int_{0}^{2\pi} \int_{0}^{1} r \, dr \, d\theta

Integrate with respect to r:

\int_{0}^{1} r \, dr = \frac{r^2}{2} \Big|_0^1 = \frac{1}{2}

V = h \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \frac{h}{2} \int_{0}^{2\pi} d\theta

Integrate with respect to θ:

\int_{0}^{2\pi} d\theta = 2\pi

V = \frac{h}{2} \cdot 2\pi = \pi h