- Given `F^'(x) = f(x)` then `int f(x)dx = F(x) + C`
Given `F_x(x,y) = f(x,y)` then `int f(x,y) dx = F(x,y) + C(y) ` - `int_(h_1(y))^(h_2(y))f_x(x,y)dx = f(x,y)|_(h_1(y))^(h_2(y)) = f(h_2(y),y) - f(h_1(y),y)`
`int_(g_1(x))^(g_2(x))f_y(x,y)dy = f(x,y)|_(g_1(x))^(g_2(x)) = f(x,g_2(x)) - f(x,g_1(x))` - iterated integrals:
`int_a^b[int_c^df(x,y) dy]dx = int_a^bint_c^df(x,y) dy dx`
`int_a^b[int_(g_1(x))^(g_2(x))f(x,y) dy]dx = int_a^bint_(g_1(x))^(g_2(x))f(x,y) dy dx`
`int_c^d[int_(h_1(y))^(h_2(y))f(x,y) dx]dy = int_c^dint_(h_1(y))^(h_2(y))f(x,y) dx dy` - developing iterated integrals as an area of the plane between two curves
- The way we used to do it: `int_a^b(g_2(x) - g_1(x))dx`
- The Fundamental Theorem of Calculus: `int_a^bf(x)dx = F(b) - F(a)`
- but if we form `int_(g_1(x))^(g_2(x))dy`, then by the FTC
`int_(g_1(x))^(g_2(x))dy = y|_(g_1(x))^(g_2(x)) = g_2(x) - g_1(x)`, considering x a constant - replacing `g_2(x) - g_1(x)` in the old way with the
equivalent integral that we just created gives:
`int_a^bint_(g_1(x))^(g_2(x))dy dx = int_a^b(g_2(x) - g_1(x))dx`
- If `R` is defined by `a <= x <= b` and `g_1(x) <= y <= g_2(x)`,
where `g_1` and `g_2` are continuous on `[a,b]`, then the area
of `R` is given by
`A = int_a^bint_(g_1(x))^(g_2(x))dydx` - If `R` is defined by `c <= y <= d` and `h_1(y) <= x <= h_2(y)`,
where `h_1` and `h_2` are continuous on `[c,d]`, then the area
of `R` is given by
`A = int_c^dint_(h_1(y))^(h_2(y))dxdy`
- Definition of Double Integral:
If `f` is defined on a closed, bounded region `R` in the `xy`-plane, then the double integral of `f` over `R` is given by
`int_RintF(x,y)dA = lim_(||Delta||rarr0)sum_(i=1)^nf(x_i,y_i)DeltaA_i`, provided the limit exists. If the limit exists, then `f` is integrable over `R`. - Volume of a Solid:
`V = int_Rintf(x,y)dA` - Usual properties
- Evaluation: change to an iterated integral. Thus, for a plane region `R`
- if `R` is defined by `a<=x<=b` and `g_1(x)<=y<=g_2(x)`, where `g_1` and `g_2` are continuous on `[a,b]`, then `int_Rintf(x,y)dA = int_a^bint_(g_1(x))^(g_2(x))f(x,y)dy dx`
- if `R` is defined by `c<=y<=d` and `h_1(y)<=x<=h_2(y)`, where `h_1` and `h_2` are continuous on `[c,d]`, then `int_Rintf(x,y)dA = int_c^dint_(h_1(y))^(h_2(y))f(x,y)dx dy`
- Average value of a function over a region:
average value of `R` is `(1/A)int_RintF(x,y)dA`, where `A` is the area of `R`.
- polar sectors
`R = {(r,theta): r_1 <= r <= r_2`, ` theta_1 <= theta <= theta_2}` - Area of polar sector from `r_1` to `r_2` and rotation `Deltatheta` is
`DeltaA_i = r_iDeltar_iDeltatheta_i` where `r_i=(r_2+r_1)/2` and `Deltar_i=r_2-r_1` - `f(x,y) = f(rcostheta,rsintheta)`
- volume of solid above `(r_i,theta_i)` is approximately `f(r_1costheta_i,r_isintheta_i)r_iDeltar_iDeltatheta_i`
- `int_Rint f(x,y)dA ~~ sum_(i=1)^nf(r_icostheta_i,r_isintheta_i)r_iDeltar_iDeltatheta_i`
- For regions where `theta` varies between constants `alpha` and `beta`
and where `r` varies between `g_1(theta) \quad and \quad g_2(theta)` for any given `theta`
`int_Rintf(x,y)dA = int_alpha^betaint_(g_1(theta))^(g_2(theta))f(rcostheta,rsintheta)*r\quaddr d theta`
Note that area of `R = int_RintdA = int_alpha^beta int_(g_1(theta))^(g_2(theta)) r dr d theta` and that there is a pencast showing such a problem along with a web page that gives the screen shots of using a calculator to evaluate the expressions that we produce in the pencast. - For regions where `r` varies between constants `a` and `b`
and where `theta` varies between `h_1(r)\quad and \quad h_2(r)` for any given `r`
`int_Rintf(x,y)dA = int_a^bint_(h_1(r))^(h_2(r))f(rcostheta,rsintheta)*r \quad d theta d r`
- Let `rho` be a continuous density function over a lamina (a plane region `R`,
then the mass of the lamina is given by
`m = int_Rint rho(x,y)dA` - Moments of mass with respect to the x and y axes:
`M_x = int_Rint y rho(x,y)dA` and `M_y = int_Rint x rho(x,y)dA` - Center of mass where `m` is the mass of the lamina, is at `(bar x,bar y) = ((M_y)/m,(M_x)/m)`
- The term centroid is used `(bar x,bar y)`for a simple plane region.
- Moments of Inertia (measure of resistence to change in a rotational motion)
about the x and y axes:
`I_x = int_Rint y^2 rho(x,y)dA` and `I_y = int_Rint x^2 rho(x,y)dA` - Polar moment of inertia as the sum of `I_x` and `I_y`
`I_0 = int_Rint(x^2+y^2) rho(x,y)dA = int_Rint r^2 rho(x,y)dA` - Radius of Gyration, `barbarr` is defined as
`barbarr=sqrt(I/m)` where `I` is the moment of intertia and `m` is the mass.
2-D Length on x-axis `int_a^b dx ` Arc length in xy-plane `int_a^b ds = int_a^b sqrt(1+[f^'(x)]^2) " "dx ` 3-D Area in xy-plane `int_Rint dA` Surface area in space `int_Rint dS = int_Rint (sqrt(1+[f_x(x,y)]^2 +[f_y(x,y)]^2)) " "dA` - Picture of one area:
`u=(:Deltax,0,f_x(x,y)Deltax:)` and `v=(:0,Deltay,f_y(x,y)Deltay:)`
To get the area of the parallelogram we find `||uxxv||`
`uxxv = |(i,j,k),(Deltax,0,f_x(x,y)Deltax),(0,Deltay,f_y(x,y)Deltay)|`
`uxxv = (-f_x(x,y)DeltaxDeltay)bbi - (f_y(x,y)DeltaxDeltay)bbj+ (DeltaxDeltay)bbk`
`uxxv = (-f_x(x,y)bbi - f_y(x,y)bbj + 1bbk)DeltaA`
`||uxxv|| = (sqrt( [f_x(x,y)]^2 + [f_y(x,y)]^2 + 1 ) )DeltaA`
- definition: For a solid region `Q`
`intint_Qintf(x,y,z)dV = lim_(||Delta||rarr0)sum_(i=1)^nf(x_i,y_i,z_i)DeltaV_i`
provided the limit exists. - Then the volume of a solid `Q` is given by
`intint_QintdV` - How do we evaluate a triple integral? We use iterated integrals when the problem lends itself to that.
- So if our region `Q` is defined by
`a<=x<=b` and `h_1(x)<=y<=h_2(x)` and `g_1(x,y)<=z<=g_2(x,y)` with all continuous functions, then
`intint_Qintf(x,y,z)dV = int_a^b int_(h_1(x))^(h_2(x))int_(g_1(x,y))^(g_2(x,y))f(x,y,z) dz dy dx` - Note the order of integrations, and that there are other possible arrangements.
- Determining the limits of integration.
- Center of Mass and Moments of Inertia
All of this assumes a density function, `rho`, for the solid. `m = intint_Qint rho(x,y,z) dV` Mass of the solid `M_(yz) = intint_Qint x rho(x,y,z) dV` First moment about the yz-plane `M_(xz) = intint_Qint y rho(x,y,z) dV` First moment about the xz-plane `M_(xy) = intint_Qint z rho(x,y,z) dV` First moment about the xy-plane `(bar x","bar y","bar z) = ((M_(yz))/m"," (M_(xz))/m","(M_(xy))/m)` coordinates of the center of mass `I_x = intint_Qint (y^2+z^2) rho(x,y,z) dV` Moment of intertia about the x-axis `I_y = intint_Qint (x^2+z^2) rho(x,y,z) dV` Moment of intertia about the y-axis `I_z = intint_Qint (x^2+y^2) rho(x,y,z) dV` Moment of intertia about the z-axis Alternative values that ease calculations: `I_(xy) = intint_Qint z^2 rho(x,y,z) dV` `I_(xz) = intint_Qint y^2 rho(x,y,z) dV` `I_(yz) = intint_Qint x^2 rho(x,y,z) dV` `I_x = I_(xz)+I_(xy)` ; `I_y = I_(yz)+I_(xy)` ; `I_z = I_(yz)+I_(xz)`
- If we have a function `f(x,y,z)` and we want to find the
`intint_Qintf(x,y,z)dV`
we can move to cylindrical coordinates `(r,theta,z)` remembering that
`x=rcos theta` and `y=rsin theta` and `z=z` - If our solid can be described by
`{(r,theta,z)\quad |\quad theta_1<=theta<=theta_2 ` and ` g_1(theta) <= r <= g_2(theta) ` and `h_1(r cos theta, r sin theta) <= z <= h_2(r cos theta, r sin theta) }`
then `intint_Qint f(x,y,z) dV = int_(theta_1)^(theta_2)int_(g_1(theta))^(g_2(theta))int_(h_1(r cos theta, r sin theta))^(h_2(r cos theta, r sin theta))f(r cos theta, r sin theta, z)*r" "dz dr d theta` - That means that for such a region, the volume of the region will be `intint_Qintf(x,y,z)dV = int_(theta_1)^(theta_2)int_(g_1(theta))^(g_2(theta))int_(h_1(r cos theta, r sin theta))^(h_2(r cos theta, r sin theta))r" "dz dr d theta`
- Note similar changes to iterated integrals for r-simple regions.
- Then, for completeness, remembering that if a point has rectangular coordinates `(x,y,z)`
and has equivalent spherical coodintes `(rho, theta, phi)` we have the conversions:
` x = rho sin phi cos theta`
` y = rho sin phi sin theta` and
` z= rho cos phi` - And it turns out that
`intint_Qint f(x,y,z)dV = `
`int_(theta_1)^(theta_2)int_(phi_1)^(phi_2)int_(rho_1)^(rho_2) f(rho sin phi cos theta, rho sin phi sin theta, rho cos phi)*rho^2*sin phi" "d rho d phi d theta`
- Back in the single variable case, if we have a function `g` such that `x=g(t)` and
for specific values `a` and `b` we have associated values `c` and `d` such that
`g(c)=a` and `g(d)=b`, then, in the case where we might want to use a change of variables,
we have `dx=g^'(t)dt` and
`int_a^bf(x)dx = int_c^df(g(t))g^'(t)dt`. - For double integrals we have:
Definition:
If `x=g(u,v)` and `y=h(u,v)`, then the Jacobian of `x` and `y` with respect to `u` and `v`, is denoted by `(del(x,y))/(del(u,v))`, is
`(del(x,y))/(del(u,v)) = |[(del x)/(del u),(del x)/(del v)],[(del y)/(del u),(del y)/(del v)]| = (del x)/(del u) (del y)/(del v) - (del y)/(del u) (del x)/(del v) `
(Note that the vertical lines signify a determinant.) - Note that the symbol for the Jacobian, namely, `(del(x,y))/(del(u,v))`, is not surrounded by vertical lines. If we use `|(del(x,y))/(del(u,v))|` we are signifying the absolute value of the Jacobian.
- Changing variables from `(x,y)` over a region `R` to `(u,v)` over a region `S`
where `x=g(u,v)` and `y=h(u,v)` then
`int_Rintf(x,y)dx dy = int_Sintf(g(u,v),h(u,v))*|(del(x,y))/(del(u,v))|*du dv` - Jacobian for Rectangular to Polar Conversion
`x = r\quad cos\quad theta` and `y = r\quad sin\quad theta`
`(del(x,y))/(del(u,v)) = |[(del x)/(del u),(del x)/(del v)],[(del y)/(del u),(del y)/(del v)]| = |[cos theta,-r sin theta],[sin theta, rcos theta]|= rcos^2theta + rsin^2theta = r`
- See the First example page for changing variables and using the Jacobean.
- See the Second example page for changing variables and using the Jacobean.
- Watch and listen to Part I of a problem best done on Chrome
- Watch and listen to Part II of a problem also best done on Chrome