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25 Cards in this Set
- Front
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R is a region in the x,y plane bounded by: x=a, x=b, y=g(x), y=h(x). Find the volume of the region. |
v = ∫ from a to b ∫ from g(x) to h(x) [f(x,y)dy]dx h(x) = top function g(x) = bottom function a to b = range of x values |
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R is a region in the x,y plane bounded by y=c, y=d, x=m(y), x=n(y). Find the volume of the region. |
v = ∫ from c to d ∫ from m(y) to n(y) [f(x,y)dx]dy n(y) = right function m(y) = left function c to d = range of y values |
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How do you tell which order, dydx or dxdy to use in a double integration? |
Sketch the region (2 dimensional), if the graph changes what is on top or bottom: use dxdy if the graph changes what is on left or right: use dydx. Whenever a change occurs, we must use a separate double integral. Find the intersections of the functions by setting the functions equal to each other. |
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Change the order of integration. |
Some functions are impossible or difficult to evaluate in one order of integration. To change the order: Sketch the functions. Find the intersections. Inner most integral bounds are taken with respect to inner partial. When you flip the order, examine which function is on the left or right, top or bottom, place that function into its place on the integral with respect to the variable you are integrating. Outer bounds are the range of x or y values, examine the graph and use the intersections to find where you enter and leave the region in the y or x axis. |
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Polar conversions |
x²+y²=r² or r=√x²+y² sinθ = y/r or y=rsinθ cosθ = x/r or x=rcosθ tanθ = y/x or θ = arctan(y/x) +π if x<0 |
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R is a bounded region, use polar coordinates to find the volume over R with the function and bounds given. |
v = ∫ from g(θ) to h(θ) ∫ from α to β [f(cosθ, rsinθ)rdr]dθ h(θ) = furthest from the origin g(θ) = closest to the origin α to β = range of polar values Note: if the origin is part of the region the bottom limit is 0. Solve: sketch the region (2 dimensional) Find intersections of regions and boundaries r integral bounds: a projection of line r through the region, where does it enter (bottom bound) where does it exit (top bound) translate this into polar form θ bounds: angle θ rotated along region, where does the region start (bottom bound) where does the region end (top bottom) translate to polar from 0 to 2π |
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Why is there an extra r in polar double integrals? |
Because of the shape of the Riemann sum. |
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When do we use polar double integration? |
When it makes our lives easier. This means functions that change often, or translate neatly into polar form. |
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double integral for total mass of given density function of a lamina δ (mass/area) |
m = ∫∫ over the region [δ(x,y)]dA dA being order of integration of the region |
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Double integral for center of mass in a lamina in the shape of R with δ(x,y) |
(∫∫xδ(x,y)dA / ∫∫δ(x,y)dA , ∫∫yδ(x,y)dA / ∫∫δ(x,y)dA) |
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Double integral for average value of f(x,y) on R |
∫∫f(x,y)dA / ∫∫ 1dA
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Double integral for moment of inertia |
Ix = ∫∫y²δ(x,y)dA : rotated lamina about x Iy = ∫∫x²δ(x,y)dA : rotated lamina about y Io = ∫∫x²+y²δ(x,y)dA : rotated lamina about origin |
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Double integral for surface area over a region |
SA = ∫∫ √fx(x,y)²+fy(x,y)²+1 dA fx and fy are partials Surface area problems use 2 graphs. One is the three dimensional representation of the region. One is the projection of that region to determine boundary limits. |
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Triple integral: Given a density function, find total mass of region R bounded by D as z= g(x,y), z=(h(x,y) and the projection of R into the x,y plane. |
∫∫ over R ∫ from h(x,y) to g(x,y) [f(x,y,x)dz]dA
g(x,y) = bounded by this top function h(x,y) = bounded by this bottom function over R = the projection of R into the x, y plane. We need 2 graphs. One for the 3D surface, one for the x,y plane projection for bounds. dA is determined by the x,y projection graph using the easiest order of integration. (does the top or bottom change? the left or right functions change?) |
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Triple integral applications: |
See card on double integral applications. Add a third integral or variable to all of them. |
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Triple integral moment of inertia |
Ix = ∫∫∫(y²+z²)p(x,y,z)dV : about x Iy = ∫∫∫(x²+z²)p(x,y,z)dV : about y Iz = ∫∫∫(x²+y²)p(x,y,z)dV : about z Io = ∫∫∫(x²+y²+z²)p(x,y,z)dV : about origin where dV is the order of integration of three variables where p is the given density function |
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Common spherical functions to evaluate in triple integrals |
common functions easy to evaluate in spherical: Sphere: x²+y²+z²=R² or p=R Cone: z=√x²+y² or ∅ = π/4 z=√3x²+3y² or ∅ = π/6 z= 0 or ∅= π/2 z= √x²/3 + y²/3 or ∅=π/3 |
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Spherical triple integrals |
∫∫∫f(psin∅cosθ, psin∅sinθ, pcos∅)p²sin∅dpd∅dθ
inner bounds: from solid closest to origin to solid furthest from origin in terms of p middle bounds: rotate your arm about the region, this is the angle from the z-axis as ∅ outer bounds: how far you go around z axis as θ note: r is always positive in the bounds. if origin is included in region, bound is 0 |
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Cylinderical triple integrals |
∫∫∫f(rcosθ,rsinθ,z)rdzdrdθ
inner bounds: function closes to origin in terms of r middle bounds: rotational arm r where you enter and exit the region outer bounds: range of θ values |
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Jacobian coordinate transformation |
∫∫f(x,y)dA = ∫∫ f(m(u,v), n(u,v))||dx/du dx/dv||dA
||dy/du dy/dv|| where u=g(x,y) ; v=h(x,y) ; x=m(u,v) ; y= n(u,v) matrix is detonated above with an absolute value as an adjustment factor. Where the integrals are evaluated as different regions.[ |
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Steps of a Jacobian transformation |
1.) solve u and v in terms of x and y as a system of equations. 2.) find partials of x and y with respect to u and v 3.) sub into matrix 4.) simplify and solve matrix 5.) change boundaries using u and v 6.) integrate |
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Jacobian matrix in cylindrical |
||cosθ -rsinθ|| ||sinθ rcosθ ||= r(cos²θ+sinθ)=r (note that this is the adjusment factor in intergration of cylindrical coordinates
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Jacobian matrix in spherical |
dx/dp dx/d∅ dx/dθ dy/dp dy/d∅ dy/dθ dz/dp dz/d∅ dz/dθ = p²sin∅ (adjustment in spherical integration) |
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Transformation for parallelogram to a square |
Ax+By=C1 Ax+By=c2 Dx+Ey=F1 Dx+Ey=F2 u=Ax+By V=Dx+Ey |
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Transformation for a ellipse to a circle |
x²/A² + y²/B² = 1 x= Au y=Bv Au²/A² + Bv²/v² = 1 or u²+v²=1 |
