Multiple integrals

For a function $f(x,y)$ defined on a region $R\subset {\mathbb{R}}^2$, the double integral is

where $dA$ is an infinitesimal area element.

• It generalizes the definite integral to two variables.
• Geometric meaning: represents the volume under $z=f(x,y)$ above region $R$.

Evaluate

Solution:

Answer: $7$.

Evaluate

Solution:

Answer: $\frac{2}{3}$.

Evaluate

where $R$ is bounded by $y=x^2,y=1,x=0,x=1$.

Solution:

Answer: $\frac{13}{20}$.

The area of a region $R$ is:

Find the area between $y=x^2$ and $y=x$ for $0\le x\le 1$.

Solution:

Answer: $\frac{1}{6}$.

Evaluate

Solution:
Region: $0\le x\le 1,0\le y\le \sqrt{x}$.
Equivalent: $0\le y\le 1,y^2\le x\le 1$.

So,

Which is simpler to handle than the original.

Let $f,g$ be integrable on $R$.

1. Linearity:

2. Additivity over regions:

if $R=R_1\cup R_2$, disjoint.

3. Positivity: If $f\ge 0$, then ${\iint }_{R}fdA\ge 0$.

4. Comparison: If $f\le g$, then

5. Zero function: If $f=0$, integral is $0$.
6. Absolute value inequality:

7. Mean Value Theorem: If $f$ continuous, then

for some $(x_0,y_0)\in R$.

Compute

Solution:

First term:

Second term:

So total:

Find the area of a circle of radius $a$.

Solution:

Evaluate

Solution:

Find the area between $r=1$ and $r=2$, with $0\le \theta \le \pi$.

Solution:

Compute

Solution:

If the region $R$ is symmetric about an axis or the origin, and the integrand has certain parity properties (odd/even), then the integral may simplify dramatically.

Evaluate

Solution:
The region is the unit disk, symmetric about the y-axis.
The function $f(x,y)=x$ is odd in $x$.
By symmetry, the integral vanishes:

Evaluate

Solution:
Region: square symmetric in both x and y.
Function: $f(x,y)=x^{2}y$.

• In $y$, it is odd ($f(x,-y)=-f(x,y)$).
• Region symmetric about x-axis.
  Therefore:

Evaluate

Solution:
Both terms are even in $x$ and $y$.
So we can compute only in the first quadrant and multiply by 4:

Inner integral:

Outer:

Multiply by 4:

If a thin lamina occupies region $R$ with density $\rho (x,y)$, its mass is

The centroid $(\overline{x},\overline{y})$ is given by

Find the centroid of the semicircular region $x^2+y^2\le a^2,y\ge 0$ with uniform density.

Solution:

• Mass:

• By symmetry, $\overline{x}=0$.

• For $\overline{y}$:

Use polar coordinates:

Divide by mass:

Centroid: $(0,\frac{4a}{3\pi })$.

Find the mass of the unit disk with density $\rho (x,y)=x^2+y^2$.

Solution:
In polar coordinates:

Compute the average value of $f(x,y)=x+y$ on $R=[0,1]\times [0,1]$.

Solution:
Average value:

Here $A(R)=1$.
So:

Answer: $1$.

For a lamina with density $\rho (x,y)$ over a region $R$, the moments of inertia measure its resistance to rotation about an axis.

• About the x-axis ($I_x$):

• About the y-axis ($I_y$):

• About the origin (Polar Moment, $I_0$): This measures resistance to twisting.

The radius of gyration is the distance from the axis at which the entire mass could be concentrated without changing the moment of inertia.

Find the moments of inertia $I_x$, $I_y$, and $I_0$ for a rectangular lamina with constant density $\rho =1$ defined by $R=[0,b]\times [0,h]$.

Solution:
First, calculate the mass $M=\rho \cdot \text{Area}=1\cdot bh=bh$.

• Moment about x-axis:

• Moment about y-axis:

• Polar Moment:

The area of the surface $z=f(x,y)$ over a region $R$ in the xy-plane is given by

Derivation Idea: We approximate a small patch of the surface with a patch of its tangent plane. The area of this tangent plane patch is larger than its shadow $dA$ in the xy-plane. The scaling factor is $\sqrt{f_x^2+f_y^2+1}$, which is the magnitude of the normal vector to the surface divided by its z-component.

The area of the surface $f(x,y,z)=c$ over a closed and bounded plane region $\mathbb{R}$ is $$ \pu{ Surface area = } \iint_{\mathbb{R}} \frac{|\nabla f|}{|\nabla f \cdot \mathbf{p}|} \dd A $$ where $\mathbf{p}$ is normal vector to $\mathbb{R}$ and $\mathrm{\nabla }f\cdot \mathbf{p}\ne 0$

Find the area of the portion of the plane $z=4-x-2y$ that lies above the rectangular region $R=[0,1]\times [0,1]$ in the xy-plane.

Solution:
Let $f(x,y)=4-x-2y$. First, we find the partial derivatives.

Now, we plug these into the surface area formula.

The integral becomes:

Since $R$ is a unit square, its area is $1$.

Answer: $\sqrt{6}$.

For a function $f(x,y,z)$ defined on a solid region $E\subset {\mathbb{R}}^3$, the triple integral is

where $dV$ is an infinitesimal volume element, $dV=dxdydz$.

• It generalizes the double integral to three variables.
• If $f=1$, the integral gives the volume of the region $E$.
• If $f$ is a density function $\rho (x,y,z)$, the integral gives the mass of the solid.

Evaluate ${\iiint }_{E}zdV$ where $E$ is the solid tetrahedron bounded by the planes $x=0$, $y=0$, $z=0$, and $x+y+z=1$.

Solution:

1. Set the bounds for z: The solid is bounded below by $z=0$ and above by the plane $z=1-x-y$.
2. Project onto the xy-plane: The projection $R$ is the triangle bounded by $x=0$, $y=0$, and $x+y=1$.
3. Set the bounds for R (Type I): For this triangle, $x$ goes from $0$ to $1$, and for each $x$, $y$ goes from $0$ to $1-x$.

Now, set up the iterated integral:

• Innermost integral (z):

• Middle integral (y):

• Outermost integral (x):

Answer: $\frac{1}{24}$.

Find the volume of the solid that lies within the cylinder $x^2+y^2=1$, below the plane $z=4$, and above the paraboloid $z=1-x^2-y^2$.

Solution:
In cylindrical coordinates:

• The cylinder is $r=1$.
• The plane is $z=4$.
• The paraboloid is $z=1-r^2$.

The bounds are:

• $z$: from $1-r^2$ to $4$.
• $r$: from $0$ to $1$.
• $\theta$: from $0$ to $2\pi$.

The volume is ${\iiint }_{E}1dV$.

• Inner integral ($z$):

• Middle integral ($r$):

• Outer integral ($\theta$):

Answer: $\frac{7\pi }{2}$.

Find the volume of the solid that lies above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=z$.

Solution:
First, convert the boundaries to spherical coordinates.

• Cone: $z=\sqrt{x^2+y^2}⟹\rho \cos \varphi =r=\rho \sin \varphi ⟹\tan \varphi =1⟹\varphi =\pi /4$.
• Sphere: $x^2+y^2+z^2=z⟹{\rho }^2=\rho \cos \varphi ⟹\rho =\cos \varphi$.

The bounds are:

• $\rho$: from $0$ to $\cos \varphi$.
• $\varphi$: from $0$ to $\pi /4$.
• $\theta$: from $0$ to $2\pi$.

Set up the volume integral:

• Inner integral ($\rho$):

• Middle integral ($\varphi$): (Use u-sub: $u=\cos \varphi ,du=-\sin \varphi d\varphi$)

• Outer integral ($\theta$):

Answer: $\frac{\pi }{8}$.

For a transformation from $uv$-coordinates to $xy$-coordinates given by $x=g(u,v)$ and $y=h(u,v)$, the Jacobian determinant is:

The Jacobian relates the area element in the two coordinate systems:

Let $R$ be a region in the $xy$-plane and $S$ be its corresponding region in the $uv$-plane under the transformation. Then:

This formula extends to triple integrals with a $3\times 3$ Jacobian matrix.

The change of variables we've already used are special cases of this general theorem.

• Polar: $x=r\cos \theta ,y=r\sin \theta$. The Jacobian is $J(r,\theta )=r$.

So $dA=|r|drd\theta =rdrd\theta$ (since $r\ge 0$).

• Cylindrical: $x=r\cos \theta ,y=r\sin \theta ,z=z$. The Jacobian is $J(r,\theta ,z)=r$.
• Spherical: $x=\rho \sin \varphi \cos \theta ,y=\rho \sin \varphi \sin \theta ,z=\rho \cos \varphi$. The Jacobian is $J(\rho ,\varphi ,\theta )={\rho }^2\sin \varphi$.

Evaluate ${\iint }_R(x-y)dA$ where $R$ is the parallelogram enclosed by the lines $x-y=0$, $x-y=2$, $x+2y=0$, and $x+2y=3$.

Solution:
The region $R$ is complex in Cartesian coordinates. We use a change of variables suggested by the boundary lines.

Let $u=x-y$ and $v=x+2y$.
This transforms the region $R$ in the $xy$-plane to a simple rectangle $S$ in the $uv$-plane defined by $0\le u\le 2$ and $0\le v\le 3$.

We need to express $x$ and $y$ in terms of $u$ and $v$ to find the Jacobian.

• From $u=x-y⟹x=u+y$.
• Substitute into $v=x+2y⟹v=(u+y)+2y⟹v-u=3y⟹y=\frac{1}{3}(v-u)$.
• Then $x=u+\frac{1}{3}(v-u)=\frac{2}{3}u+\frac{1}{3}v$.

Now, calculate the Jacobian $J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}$:

The absolute value is $|J(u,v)|=1/3$. Now, transform the integral:

Answer: $2$.