Apply double and triple integration techniques to evaluate area, volume, and related engineering problems.
A double integral extends the concept of a single integral to functions of two variables. It computes the signed volume under a surface \( z = f(x, y) \) over a region \( R \) in the \( xy \)-plane.
Divide the region \( R \) into \( n \) small rectangles of area \( \Delta A_i \). Choose a sample point \( (x_i, y_i) \) in each rectangle. Then:
| Property | Formula |
|---|---|
| Linearity (constant) | \( \iint_R k f\,dA = k \iint_R f\,dA \) |
| Linearity (sum) | \( \iint_R (f \pm g)\,dA = \iint_R f\,dA \pm \iint_R g\,dA \) |
| Additivity | If \( R = R_1 \cup R_2 \) (non-overlapping), \( \iint_R f\,dA = \iint_{R_1} f\,dA + \iint_{R_2} f\,dA \) |
| Area of region | \( A(R) = \iint_R 1\,dA \) |
| Non-negativity | If \( f \geq 0 \) on \( R \), then \( \iint_R f\,dA \geq 0 \) |
| Comparison | If \( f \leq g \) on \( R \), then \( \iint_R f\,dA \leq \iint_R g\,dA \) |
A region bounded by \( x = a \), \( x = b \), \( y = g_1(x) \), \( y = g_2(x) \) with \( g_1(x) \leq g_2(x) \):
A region bounded by \( y = c \), \( y = d \), \( x = h_1(y) \), \( x = h_2(y) \) with \( h_1(y) \leq h_2(y) \):
Double integrals are evaluated by iterated integration — integrating with respect to one variable at a time while treating the other as constant.
Problem: Evaluate \( \displaystyle\int_0^1 \int_0^2 (x + y)\,dy\,dx \)
Solution:
Step 1: Inner integral (w.r.t. \( y \), treating \( x \) as constant):
\[ \int_0^2 (x + y)\,dy = \left[xy + \frac{y^2}{2}\right]_0^2 = 2x + 2 \]Step 2: Outer integral (w.r.t. \( x \)):
\[ \int_0^1 (2x + 2)\,dx = \left[x^2 + 2x\right]_0^1 = 1 + 2 = 3 \]Therefore, \( \displaystyle\int_0^1 \int_0^2 (x + y)\,dy\,dx = 3 \)
Problem: Evaluate \( \displaystyle\int_0^1 \int_0^x (x^2 + y^2)\,dy\,dx \)
Solution:
Step 1: Inner integral:
\[ \int_0^x (x^2 + y^2)\,dy = \left[x^2 y + \frac{y^3}{3}\right]_0^x = x^3 + \frac{x^3}{3} = \frac{4x^3}{3} \]Step 2: Outer integral:
\[ \int_0^1 \frac{4x^3}{3}\,dx = \frac{4}{3}\cdot\frac{x^4}{4}\bigg|_0^1 = \frac{1}{3} \]Therefore, \( \displaystyle\int_0^1 \int_0^x (x^2 + y^2)\,dy\,dx = \frac{1}{3} \)
Problem: Evaluate \( \displaystyle\int_0^1 \int_0^1 e^{x+y}\,dx\,dy \)
Solution:
Step 1: Inner integral (w.r.t. \( x \)):
\[ \int_0^1 e^{x+y}\,dx = e^y \int_0^1 e^x\,dx = e^y[e^x]_0^1 = e^y(e - 1) \]Step 2: Outer integral (w.r.t. \( y \)):
\[ \int_0^1 e^y(e - 1)\,dy = (e - 1)[e^y]_0^1 = (e - 1)^2 \]Therefore, \( \displaystyle\int_0^1 \int_0^1 e^{x+y}\,dx\,dy = (e - 1)^2 \)
Problem: Evaluate \( \displaystyle\iint_R xy\,dA \) where \( R \) is the triangle with vertices \( (0,0) \), \( (1,0) \), \( (1,1) \).
Solution:
The region is bounded by \( x = 0 \) to \( x = 1 \), \( y = 0 \) to \( y = x \).
\[ \iint_R xy\,dA = \int_0^1 \int_0^x xy\,dy\,dx \]Inner integral:
\[ \int_0^x xy\,dy = x\left[\frac{y^2}{2}\right]_0^x = \frac{x^3}{2} \]Outer integral:
\[ \int_0^1 \frac{x^3}{2}\,dx = \frac{1}{2}\cdot\frac{x^4}{4}\bigg|_0^1 = \frac{1}{8} \]Therefore, \( \displaystyle\iint_R xy\,dA = \frac{1}{8} \)
Sometimes a double integral is difficult (or impossible) to evaluate in the given order. Changing the order of integration can simplify the calculation.
Problem: Change the order of integration in \( \displaystyle\int_0^1 \int_x^1 f(x, y)\,dy\,dx \).
Solution:
Step 1: The region is bounded by \( x = 0 \) to \( x = 1 \), \( y = x \) to \( y = 1 \).
Step 2: This is the triangle with vertices \( (0,0) \), \( (1,1) \), \( (0,1) \).
Step 3: Reversing the order: \( y \) goes from 0 to 1, and for each \( y \), \( x \) goes from 0 to \( y \).
\[ \int_0^1 \int_x^1 f(x, y)\,dy\,dx = \int_0^1 \int_0^y f(x, y)\,dx\,dy \]Problem: Evaluate \( \displaystyle\int_0^1 \int_x^1 e^{y^2}\,dy\,dx \) by changing the order.
Solution:
The original integral cannot be evaluated directly because \( e^{y^2} \) has no elementary antiderivative.
Step 1: Region: \( 0 \leq x \leq 1 \), \( x \leq y \leq 1 \) (triangle).
Step 2: Change order: \( 0 \leq y \leq 1 \), \( 0 \leq x \leq y \).
\[ \int_0^1 \int_x^1 e^{y^2}\,dy\,dx = \int_0^1 \int_0^y e^{y^2}\,dx\,dy \]Step 3: Inner integral (w.r.t. \( x \)):
\[ \int_0^y e^{y^2}\,dx = e^{y^2}\cdot y \]Step 4: Outer integral (w.r.t. \( y \)):
\[ \int_0^1 y e^{y^2}\,dy \]Let \( u = y^2 \), \( du = 2y\,dy \):
\[ = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}(e - 1) \]Therefore, \( \displaystyle\int_0^1 \int_x^1 e^{y^2}\,dy\,dx = \frac{e-1}{2} \)
Problem: Evaluate \( \displaystyle\int_0^a \int_y^a \frac{x}{x^2 + y^2}\,dx\,dy \) by changing the order.
Solution:
Step 1: Region: \( 0 \leq y \leq a \), \( y \leq x \leq a \) (triangle).
Step 2: Change order: \( 0 \leq x \leq a \), \( 0 \leq y \leq x \).
\[ \int_0^a \int_0^x \frac{x}{x^2 + y^2}\,dy\,dx \]Step 3: Inner integral:
\[ \int_0^x \frac{x}{x^2 + y^2}\,dy = \left[\tan^{-1}\frac{y}{x}\right]_0^x = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} \]Step 4: Outer integral:
\[ \int_0^a \frac{\pi}{4}\,dx = \frac{\pi a}{4} \]Therefore, \( \displaystyle\int_0^a \int_y^a \frac{x}{x^2 + y^2}\,dx\,dy = \frac{\pi a}{4} \)
When the region of integration is a circle, annulus, or sector, converting to polar coordinates simplifies the integral significantly.
The area element \( dA = dx\,dy \) transforms as:
This is derived from the Jacobian:
\[ J = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r \]Problem: Evaluate \( \displaystyle\iint_R (x^2 + y^2)\,dA \) where \( R \) is the disk \( x^2 + y^2 \leq 4 \).
Solution:
In polar coordinates: \( x^2 + y^2 = r^2 \), \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \)
\[ \iint_R (x^2 + y^2)\,dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r\,dr\,d\theta \]Inner integral:
\[ \int_0^2 r^3\,dr = \left[\frac{r^4}{4}\right]_0^2 = 4 \]Outer integral:
\[ \int_0^{2\pi} 4\,d\theta = 8\pi \]Therefore, \( \displaystyle\iint_R (x^2 + y^2)\,dA = 8\pi \)
Problem: Evaluate \( \displaystyle\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)}\,dx\,dy \).
Solution:
Convert to polar coordinates: \( x^2 + y^2 = r^2 \), \( 0 \leq r < \infty \), \( 0 \leq \theta \leq 2\pi \)
\[ = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r\,dr\,d\theta \]Let \( u = r^2 \), \( du = 2r\,dr \):
\[ \int_0^{\infty} e^{-r^2} r\,dr = \frac{1}{2}\int_0^{\infty} e^{-u}\,du = \frac{1}{2} \] \[ = \int_0^{2\pi} \frac{1}{2}\,d\theta = \pi \]Therefore, the integral equals \( \pi \).
Note: This gives \( \displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \) (a famous result).
Problem: Evaluate \( \displaystyle\iint_R \sqrt{x^2 + y^2}\,dA \) where \( R \) is the annulus \( 1 \leq x^2 + y^2 \leq 4 \).
Solution:
In polar coordinates: \( \sqrt{x^2 + y^2} = r \), \( 1 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \)
\[ \iint_R \sqrt{x^2 + y^2}\,dA = \int_0^{2\pi} \int_1^2 r \cdot r\,dr\,d\theta = \int_0^{2\pi} \int_1^2 r^2\,dr\,d\theta \]Inner: \( \displaystyle\int_1^2 r^2\,dr = \frac{8 - 1}{3} = \frac{7}{3} \)
Outer: \( \displaystyle\int_0^{2\pi} \frac{7}{3}\,d\theta = \frac{14\pi}{3} \)
Therefore, \( \displaystyle\iint_R \sqrt{x^2 + y^2}\,dA = \frac{14\pi}{3} \)
A triple integral extends integration to functions of three variables over a solid region \( V \) in three-dimensional space.
Triple integrals are evaluated as three iterated integrals. The order of integration can be any of the six possible orders (e.g., \( dz\,dy\,dx \), \( dx\,dy\,dz \), etc.).
Bounded below by \( z = h_1(x, y) \) and above by \( z = h_2(x, y) \):
Bounded by \( y = g_1(x, z) \) and \( y = g_2(x, z) \).
Bounded by \( x = f_1(y, z) \) and \( x = f_2(y, z) \).
Problem: Evaluate \( \displaystyle\int_0^1 \int_0^1 \int_0^1 (x + y + z)\,dz\,dy\,dx \)
Solution:
Step 1: Inner (w.r.t. \( z \)):
\[ \int_0^1 (x + y + z)\,dz = \left[xz + yz + \frac{z^2}{2}\right]_0^1 = x + y + \frac{1}{2} \]Step 2: Middle (w.r.t. \( y \)):
\[ \int_0^1 \left(x + y + \frac{1}{2}\right)dy = \left[xy + \frac{y^2}{2} + \frac{y}{2}\right]_0^1 = x + \frac{1}{2} + \frac{1}{2} = x + 1 \]Step 3: Outer (w.r.t. \( x \)):
\[ \int_0^1 (x + 1)\,dx = \left[\frac{x^2}{2} + x\right]_0^1 = \frac{1}{2} + 1 = \frac{3}{2} \]Therefore, the integral equals \( \frac{3}{2} \).
Problem: Find the volume of the cube \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 \), \( 0 \leq z \leq 1 \).
Solution:
\[ V = \iiint_V dV = \int_0^1 \int_0^1 \int_0^1 dz\,dy\,dx = \int_0^1 \int_0^1 1\,dy\,dx = \int_0^1 1\,dx = 1 \]Volume = 1 cubic unit.
Problem: Find the volume of the tetrahedron bounded by the planes \( x = 0 \), \( y = 0 \), \( z = 0 \), and \( x + y + z = 1 \).
Solution:
The limits are: \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 - x \), \( 0 \leq z \leq 1 - x - y \)
\[ V = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} dz\,dy\,dx \]Step 1: Inner (w.r.t. \( z \)): \( z\big|_0^{1-x-y} = 1 - x - y \)
Step 2: Middle (w.r.t. \( y \)):
\[ \int_0^{1-x} (1 - x - y)\,dy = \left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x} = (1-x)^2 - \frac{(1-x)^2}{2} = \frac{(1-x)^2}{2} \]Step 3: Outer (w.r.t. \( x \)):
\[ \int_0^1 \frac{(1-x)^2}{2}\,dx = \frac{1}{2}\left[-\frac{(1-x)^3}{3}\right]_0^1 = \frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} \]Volume = \( \frac{1}{6} \) cubic unit.
Problem: Evaluate \( \displaystyle\iiint_V (x^2 + y^2)\,dV \) where \( V \) is the cylinder \( x^2 + y^2 \leq 1 \), \( 0 \leq z \leq 2 \).
Solution:
Using cylindrical coordinates: \( x = r\cos\theta \), \( y = r\sin\theta \), \( z = z \), \( dV = r\,dr\,d\theta\,dz \)
Limits: \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq 2\pi \), \( 0 \leq z \leq 2 \)
\[ \iiint_V (x^2 + y^2)\,dV = \int_0^{2\pi} \int_0^1 \int_0^2 r^2 \cdot r\,dz\,dr\,d\theta \] \[ = \int_0^{2\pi} \int_0^1 \int_0^2 r^3\,dz\,dr\,d\theta \]Inner (w.r.t. \( z \)): \( 2r^3 \)
Middle (w.r.t. \( r \)): \( \displaystyle\int_0^1 2r^3\,dr = \frac{2}{4} = \frac{1}{2} \)
Outer (w.r.t. \( \theta \)): \( \displaystyle\int_0^{2\pi} \frac{1}{2}\,d\theta = \pi \)
Therefore, the integral equals \( \pi \).
The area of a region \( R \) in the \( xy \)-plane is given by:
Problem: Find the area bounded by \( y = x^2 \) and \( y = x \).
Solution:
Step 1: Find intersection points: \( x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1 \)
Step 2: For \( 0 \leq x \leq 1 \), \( x \geq x^2 \) (line is above parabola)
\[ A = \int_0^1 \int_{x^2}^{x} dy\,dx = \int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]Area = \( \frac{1}{6} \) square units.
Problem: Find the area of the circle \( x^2 + y^2 = a^2 \) using double integrals.
Solution:
Using polar coordinates: \( 0 \leq r \leq a \), \( 0 \leq \theta \leq 2\pi \)
\[ A = \int_0^{2\pi} \int_0^a r\,dr\,d\theta = \int_0^{2\pi} \left[\frac{r^2}{2}\right]_0^a d\theta = \int_0^{2\pi} \frac{a^2}{2}\,d\theta = \frac{a^2}{2}\cdot 2\pi = \pi a^2 \]Area = \( \pi a^2 \) square units.
Problem: Find the area enclosed between \( y^2 = 4x \) and \( x^2 = 4y \).
Solution:
Step 1: Find intersections: \( y = \frac{x^2}{4} \) and \( x = \frac{y^2}{4} \)
Substituting: \( y^2 = 4x \Rightarrow \left(\frac{x^2}{4}\right)^2 = 4x \Rightarrow \frac{x^4}{16} = 4x \Rightarrow x^4 = 64x \Rightarrow x(x^3 - 64) = 0 \)
\( x = 0 \) or \( x = 4 \). Points: \( (0,0) \) and \( (4,4) \)
Step 2: For \( 0 \leq x \leq 4 \): upper curve \( y = 2\sqrt{x} \), lower curve \( y = \frac{x^2}{4} \)
\[ A = \int_0^4 \int_{x^2/4}^{2\sqrt{x}} dy\,dx = \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right)dx \] \[ = \left[\frac{4x^{3/2}}{3} - \frac{x^3}{12}\right]_0^4 = \frac{4\cdot 8}{3} - \frac{64}{12} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]Area = \( \frac{16}{3} \) square units.
Problem: Find the area of the cardioid \( r = a(1 + \cos\theta) \).
Solution:
The cardioid is symmetric about the polar axis. We integrate from \( \theta = 0 \) to \( \theta = \pi \) and double.
\[ A = 2\int_0^{\pi} \int_0^{a(1+\cos\theta)} r\,dr\,d\theta \]Inner (w.r.t. \( r \)): \( \displaystyle\int_0^{a(1+\cos\theta)} r\,dr = \frac{a^2(1+\cos\theta)^2}{2} \)
Outer (w.r.t. \( \theta \)):
\[ A = 2\cdot\frac{a^2}{2}\int_0^{\pi} (1 + \cos\theta)^2\,d\theta = a^2\int_0^{\pi} (1 + 2\cos\theta + \cos^2\theta)\,d\theta \]Using \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \):
\[ A = a^2\int_0^{\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta \] \[ = a^2\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{\pi} = a^2\cdot\frac{3\pi}{2} \]Area = \( \frac{3\pi a^2}{2} \) square units.
The volume of a solid bounded above by \( z = f(x, y) \), below by the region \( R \) in the \( xy \)-plane, is:
The volume of a solid region \( V \) in space is:
| Coordinate System | Volume Element \( dV \) |
|---|---|
| Cartesian \( (x, y, z) \) | \( dx\,dy\,dz \) |
| Cylindrical \( (r, \theta, z) \) | \( r\,dr\,d\theta\,dz \) |
| Spherical \( (\rho, \theta, \phi) \) | \( \rho^2\sin\phi\,d\rho\,d\theta\,d\phi \) |
Problem: Find the volume of the solid bounded by the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 4 \).
Solution:
The paraboloid intersects \( z = 4 \) when \( x^2 + y^2 = 4 \) (circle of radius 2).
Using polar coordinates: \( 0 \leq r \leq 2 \), \( 0 \leq \theta \leq 2\pi \), height = \( 4 - r^2 \)
\[ V = \int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta \]Inner: \( \displaystyle\int_0^2 (4r - r^3)\,dr = \left[2r^2 - \frac{r^4}{4}\right]_0^2 = 8 - 4 = 4 \)
Outer: \( \displaystyle\int_0^{2\pi} 4\,d\theta = 8\pi \)
Volume = \( 8\pi \) cubic units.
Problem: Find the volume of the sphere \( x^2 + y^2 + z^2 = a^2 \).
Solution:
Using spherical coordinates: \( 0 \leq \rho \leq a \), \( 0 \leq \theta \leq 2\pi \), \( 0 \leq \phi \leq \pi \)
\[ V = \int_0^{2\pi} \int_0^{\pi} \int_0^a \rho^2\sin\phi\,d\rho\,d\phi\,d\theta \]Innermost (w.r.t. \( \rho \)): \( \displaystyle\int_0^a \rho^2\,d\rho = \frac{a^3}{3} \)
Middle (w.r.t. \( \phi \)): \( \displaystyle\int_0^{\pi} \sin\phi\,d\phi = [-\cos\phi]_0^{\pi} = 2 \)
Outer (w.r.t. \( \theta \)): \( \displaystyle\int_0^{2\pi} d\theta = 2\pi \)
\[ V = \frac{a^3}{3}\cdot 2\cdot 2\pi = \frac{4\pi a^3}{3} \]Volume = \( \frac{4\pi a^3}{3} \) cubic units.
Problem: Find the volume between \( z = x^2 + y^2 \) and \( z = 2 - x^2 - y^2 \).
Solution:
Step 1: Find intersection: \( x^2 + y^2 = 2 - x^2 - y^2 \Rightarrow 2(x^2 + y^2) = 2 \Rightarrow x^2 + y^2 = 1 \)
Step 2: In polar coordinates: \( 0 \leq r \leq 1 \), \( 0 \leq \theta \leq 2\pi \), height = \( (2 - r^2) - r^2 = 2 - 2r^2 \)
\[ V = \int_0^{2\pi} \int_0^1 (2 - 2r^2)\,r\,dr\,d\theta \]Inner: \( \displaystyle\int_0^1 (2r - 2r^3)\,dr = \left[r^2 - \frac{r^4}{2}\right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} \)
Outer: \( \displaystyle\int_0^{2\pi} \frac{1}{2}\,d\theta = \pi \)
Volume = \( \pi \) cubic units.
Problem: Find the volume of the cone bounded by \( z = \sqrt{x^2 + y^2} \) and \( z = h \).
Solution:
In cylindrical coordinates: \( 0 \leq r \leq h \), \( 0 \leq \theta \leq 2\pi \), \( r \leq z \leq h \)
\[ V = \int_0^{2\pi} \int_0^h \int_r^h r\,dz\,dr\,d\theta \]Innermost (w.r.t. \( z \)): \( r(h - r) \)
Middle (w.r.t. \( r \)): \( \displaystyle\int_0^h (hr - r^2)\,dr = \frac{h^3}{2} - \frac{h^3}{3} = \frac{h^3}{6} \)
Outer (w.r.t. \( \theta \)): \( \displaystyle\int_0^{2\pi} \frac{h^3}{6}\,d\theta = \frac{\pi h^3}{3} \)
Volume = \( \frac{\pi h^3}{3} \) cubic units.
For a lamina (thin plate) with density \( \rho(x, y) \):
For a lamina about the \( x \)-axis, \( y \)-axis, and origin:
For a solid with density \( \rho(x, y, z) \):
Problem: Find the centre of mass of a triangular lamina with vertices \( (0,0) \), \( (1,0) \), \( (0,1) \) and uniform density \( \rho = 1 \).
Solution:
Region: \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 - x \)
Mass:
\[ m = \int_0^1 \int_0^{1-x} dy\,dx = \int_0^1 (1-x)\,dx = \frac{1}{2} \]\( \bar{x} \):
\[ \bar{x} = \frac{1}{m}\int_0^1 \int_0^{1-x} x\,dy\,dx = 2\int_0^1 x(1-x)\,dx = 2\left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = 2\cdot\frac{1}{6} = \frac{1}{3} \]\( \bar{y} \):
\[ \bar{y} = \frac{1}{m}\int_0^1 \int_0^{1-x} y\,dy\,dx = 2\int_0^1 \frac{(1-x)^2}{2}\,dx = \int_0^1 (1-x)^2\,dx = \frac{1}{3} \]Centre of mass = \( \left(\frac{1}{3}, \frac{1}{3}\right) \)
Problem: Find the moment of inertia about the \( z \)-axis for the solid cylinder \( x^2 + y^2 \leq a^2 \), \( 0 \leq z \leq h \), with density \( \rho = 1 \).
Solution:
Using cylindrical coordinates:
\[ I_z = \iiint_V (x^2 + y^2)\,dV = \int_0^{2\pi} \int_0^a \int_0^h r^2 \cdot r\,dz\,dr\,d\theta \] \[ = \int_0^{2\pi} \int_0^a \int_0^h r^3\,dz\,dr\,d\theta \]Inner: \( hr^3 \)
Middle: \( \displaystyle h\int_0^a r^3\,dr = h\cdot\frac{a^4}{4} \)
Outer: \( \displaystyle\int_0^{2\pi} \frac{ha^4}{4}\,d\theta = \frac{\pi h a^4}{2} \)
\( I_z = \frac{\pi h a^4}{2} \)
Problem: Find the mass of the solid bounded by \( x = 0 \), \( y = 0 \), \( z = 0 \), \( x + y + z = 1 \) with density \( \rho = x \).
Solution:
Limits: \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 - x \), \( 0 \leq z \leq 1 - x - y \)
\[ m = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} x\,dz\,dy\,dx \]Inner (w.r.t. \( z \)): \( x(1 - x - y) \)
Middle (w.r.t. \( y \)): \( \displaystyle\int_0^{1-x} x(1-x-y)\,dy = x\left[(1-x)y - \frac{y^2}{2}\right]_0^{1-x} = \frac{x(1-x)^2}{2} \)
Outer (w.r.t. \( x \)): \( \displaystyle\int_0^1 \frac{x(1-x)^2}{2}\,dx \)
Let \( u = 1-x \), \( du = -dx \): \( \displaystyle\frac{1}{2}\int_0^1 (1-u)u^2\,du = \frac{1}{2}\left[\frac{u^3}{3} - \frac{u^4}{4}\right]_0^1 = \frac{1}{2}\cdot\frac{1}{12} = \frac{1}{24} \)
Mass = \( \frac{1}{24} \) units.
| Topic | Key Formula / Concept |
|---|---|
| Double Integral | \( \iint_R f(x,y)\,dA \) |
| Type I Region | \( \int_a^b \int_{g_1(x)}^{g_2(x)} f\,dy\,dx \) |
| Type II Region | \( \int_c^d \int_{h_1(y)}^{h_2(y)} f\,dx\,dy \) |
| Change of Order | \( \int_a^b \int_{g_1(x)}^{g_2(x)} f\,dy\,dx = \int_c^d \int_{h_1(y)}^{h_2(y)} f\,dx\,dy \) |
| Polar Coordinates | \( x = r\cos\theta \), \( y = r\sin\theta \), \( dA = r\,dr\,d\theta \) |
| Double Integral in Polar | \( \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta \) |
| Triple Integral | \( \iiint_V f\,dV \) |
| Cylindrical Coordinates | \( dV = r\,dr\,d\theta\,dz \) |
| Spherical Coordinates | \( dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi \) |
| Area | \( A = \iint_R dA \) |
| Volume (Double) | \( V = \iint_R f(x,y)\,dA \) |
| Volume (Triple) | \( V = \iiint_V dV \) |
| Mass | \( m = \iint_R \rho\,dA \) or \( \iiint_V \rho\,dV \) |
| Centre of Mass | \( \bar{x} = \frac{M_y}{m} \), \( \bar{y} = \frac{M_x}{m} \) |
| Moment of Inertia | \( I_x = \iint_R y^2\rho\,dA \), \( I_y = \iint_R x^2\rho\,dA \) |
| System | Conversion | Volume Element |
|---|---|---|
| Polar | \( x = r\cos\theta \), \( y = r\sin\theta \) | \( dA = r\,dr\,d\theta \) |
| Cylindrical | \( x = r\cos\theta \), \( y = r\sin\theta \), \( z = z \) | \( dV = r\,dr\,d\theta\,dz \) |
| Spherical | \( x = \rho\sin\phi\cos\theta \), \( y = \rho\sin\phi\sin\theta \), \( z = \rho\cos\phi \) | \( dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi \) |
Evaluate \( \displaystyle\int_0^1 \int_0^2 (x^2 + y^2)\,dy\,dx \)
Change the order of integration in \( \displaystyle\int_0^1 \int_{x^2}^x f(x, y)\,dy\,dx \) and evaluate.
Evaluate \( \displaystyle\iint_R e^{-(x^2+y^2)}\,dA \) where \( R \) is the disk \( x^2 + y^2 \leq a^2 \).
Evaluate \( \displaystyle\int_0^1 \int_0^{1-x} \int_0^{1-x-y} dz\,dy\,dx \)
Find the area bounded by \( y = x^2 \) and \( y = 2x - x^2 \).
Find the volume of the solid bounded by \( z = 4 - x^2 - y^2 \) and \( z = 0 \).
Find the volume of the sphere \( x^2 + y^2 + z^2 = 9 \) using triple integrals.
Find the centre of mass of a lamina with density \( \rho = 1 \) bounded by \( y = x^2 \) and \( y = 1 \).
Answer = \( \frac{10}{3} \)
Original region: \( 0 \leq x \leq 1 \), \( x^2 \leq y \leq x \) (region between \( y = x^2 \) and \( y = x \)).
Change order: \( 0 \leq y \leq 1 \), \( y \leq x \leq \sqrt{y} \)
\[ \int_0^1 \int_{x^2}^x f\,dy\,dx = \int_0^1 \int_y^{\sqrt{y}} f\,dx\,dy \]If \( f(x,y) = 1 \):
\[ \int_0^1 \int_y^{\sqrt{y}} dx\,dy = \int_0^1 (\sqrt{y} - y)\,dy = \left[\frac{2y^{3/2}}{3} - \frac{y^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2} = \frac{1}{6} \]Answer = \( \frac{1}{6} \)
In polar coordinates: \( x^2 + y^2 = r^2 \), \( 0 \leq r \leq a \), \( 0 \leq \theta \leq 2\pi \)
\[ \iint_R e^{-(x^2+y^2)}\,dA = \int_0^{2\pi} \int_0^a e^{-r^2} r\,dr\,d\theta \]Let \( u = r^2 \), \( du = 2r\,dr \):
\[ \int_0^a e^{-r^2} r\,dr = \frac{1}{2}\int_0^{a^2} e^{-u}\,du = \frac{1}{2}(1 - e^{-a^2}) \] \[ = \int_0^{2\pi} \frac{1}{2}(1 - e^{-a^2})\,d\theta = \pi(1 - e^{-a^2}) \]Answer = \( \pi(1 - e^{-a^2}) \)
Inner: \( 1 - x - y \)
Middle: \( \displaystyle\int_0^{1-x} (1-x-y)\,dy = \frac{(1-x)^2}{2} \)
Outer: \( \displaystyle\int_0^1 \frac{(1-x)^2}{2}\,dx = \frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} \)
Answer = \( \frac{1}{6} \)
Intersection: \( x^2 = 2x - x^2 \Rightarrow 2x^2 - 2x = 0 \Rightarrow 2x(x - 1) = 0 \Rightarrow x = 0, 1 \)
\[ A = \int_0^1 \int_{x^2}^{2x-x^2} dy\,dx = \int_0^1 (2x - 2x^2)\,dx = \left[x^2 - \frac{2x^3}{3}\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3} \]Area = \( \frac{1}{3} \) square units
Intersection with \( z = 0 \): \( x^2 + y^2 = 4 \) (circle radius 2)
Using polar coordinates:
\[ V = \int_0^{2\pi} \int_0^2 (4 - r^2)r\,dr\,d\theta \] \[ = \int_0^{2\pi} \left[2r^2 - \frac{r^4}{4}\right]_0^2 d\theta = \int_0^{2\pi} (8 - 4)\,d\theta = 8\pi \]Volume = \( 8\pi \) cubic units
Using spherical coordinates: \( 0 \leq \rho \leq 3 \), \( 0 \leq \theta \leq 2\pi \), \( 0 \leq \phi \leq \pi \)
\[ V = \int_0^{2\pi} \int_0^{\pi} \int_0^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta \] \[ = \frac{27}{3}\cdot 2\cdot 2\pi = 36\pi \]Volume = \( 36\pi \) cubic units
Region: \( -1 \leq x \leq 1 \), \( x^2 \leq y \leq 1 \)
Mass: \( m = \int_{-1}^1 \int_{x^2}^1 dy\,dx = \int_{-1}^1 (1 - x^2)\,dx = \left[x - \frac{x^3}{3}\right]_{-1}^1 = \frac{4}{3} \)
\( \bar{x} = 0 \)** (by symmetry)
\( \bar{y} \):
\[ \bar{y} = \frac{1}{m}\int_{-1}^1 \int_{x^2}^1 y\,dy\,dx = \frac{3}{4}\int_{-1}^1 \frac{1 - x^4}{2}\,dx = \frac{3}{8}\left[x - \frac{x^5}{5}\right]_{-1}^1 = \frac{3}{8}\cdot\frac{8}{5} = \frac{3}{5} \]Centre of mass = \( \left(0, \frac{3}{5}\right) \)
| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO5 Components | Covered in Section |
|---|---|
| Evaluate double integrals | I, II |
| Change order of integration | III |
| Use polar coordinates | IV |
| Evaluate triple integrals | V |
| Find area using double integrals | VI |
| Find volume using double/triple integrals | VII |
| Apply to engineering problems | VIII |
⭐ End of Unit V Notes ⭐
Multiple Integrals and Applications
MTH165 – Mathematics for Engineers
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