222 - Topic 14.5 - Switching order of integration -- Sage

var('x,y') f = 3/(2+y^3) plot3d(f,(x,0,4),(y,0,2))

var('x,y') @interact def _(top = input_box(y<=2), bottom = input_box(y>=sqrt(x)), xs = input_grid(1,2, default = [[0,4]],label='x range ='), ys = input_grid(1,2, default = [[0,4]],label='y range ='), x0 = slider(0,4,0.2,1)): constraints = (top, bottom) a = xs[0][0] b = xs[0][1] c = ys[0][0] d = ys[0][1] n = 100 dely = (d-c)/n G = region_plot(constraints , (x,a,b), (y,c,d), plot_points=400) for k in range(n): y0 = c+k*dely if bottom(x=x0,y=y0) and top(x=x0,y=y0): G += point((x0,y0),color='red',size=20,zorder=2) G.show()

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# Doing the problem as stated with dy first and then dx var('x,y') f = 3/(2+y^3) # We need these so that the inside integral knows everything is real valued assume(x>0) assume(sqrt(x)<2) I_y = integrate(f,y,sqrt(x),2) show(I_y) I_yx =integrate(I_y,x,0,4) show(I_yx) print "Numerically, the integral evaluates to ",I_yx.n()

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{2} \, \sqrt{3} 2^{\frac{1}{3}} \arctan\left(\frac{1}{3} \, \sqrt{3} 2^{\frac{2}{3}} \sqrt{x} - \frac{1}{3} \, \sqrt{3}\right) + \frac{1}{2} \, \sqrt{3} 2^{\frac{1}{3}} \arctan\left(\frac{1}{3} \, \sqrt{3} {\left(2 \cdot 2^{\frac{2}{3}} - 1\right)}\right) - \frac{1}{4} \cdot 2^{\frac{1}{3}} \log\left(2^{\frac{2}{3}} - 2 \cdot 2^{\frac{1}{3}} + 4\right) - \frac{1}{2} \cdot 2^{\frac{1}{3}} \log\left(\sqrt{x} + 2^{\frac{1}{3}}\right) + \frac{1}{2} \cdot 2^{\frac{1}{3}} \log\left(2^{\frac{1}{3}} + 2\right) + \frac{1}{4} \cdot 2^{\frac{1}{3}} \log\left({\left| x - 2^{\frac{1}{3}} \sqrt{x} + 2^{\frac{2}{3}} \right|}\right)$

$\newcommand{\Bold}[1]{\mathbf{#1}}-2 \, \log\left(2\right) + \frac{1}{4} \, \log\left(2^{\frac{2}{3}} - 2 \cdot 2^{\frac{1}{3}} + 4\right) + \frac{3}{4} \, \log\left(-4 \cdot 2^{\frac{2}{3}} + 8 \cdot 2^{\frac{1}{3}} + 4\right) + \log\left(2^{\frac{1}{3}} + 2\right)$

Numerically, the integral evaluates to  1.60943791243410

Numerically, the integral evaluates to  1.60943791243410

# Let's look at the region we are integrating over @interact def _(x0=slider(0,4,0.1,1.2,label='$x_0$'),y0=slider(0,2,0.1,1.8,label='$y_0$')): G = plot(2,(x,0,4),fill=True,fillcolor='green') G += plot(sqrt(x),(x,0,4),fill=True,fillcolor='white') G += text("This is the region.",[0.75,1.5]) # Vertical Strip...dy dx G += line([(x0,sqrt(x0)),(x0,1.97)],color='red',thickness=10) G += text("$y=\sqrt{x}$",[x0,sqrt(x0)-0.1]) G += text("$y=2$",[x0,2+0.1]) # Horizontal Strip...dx dy G += line([(0.03,y0),(y0^2,y0)],color='yellow',thickness=10) G += text("$x=y^2$",[y0^2+0.2,y0]) G += text("$x=0$",[0-0.1,y0]) G.show()

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# Switching the order of integration, dx and then dy I_x =integrate(f,x,0,y^2) show(I_x) I_xy =integrate(I_x,y,0,2) show(I_xy) print "Numerically, the integral evaluates to ",I_xy.n()

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{3 \, y^{2}}{y^{3} + 2}$

$\newcommand{\Bold}[1]{\mathbf{#1}}\log\left(10\right) - \log\left(2\right)$

Numerically, the integral evaluates to  1.60943791243410

Numerically, the integral evaluates to  1.60943791243410

(I_yx - I_xy).show() (I_yx - I_xy).n(digits=100)

$\newcommand{\Bold}[1]{\mathbf{#1}}-\log\left(10\right) - \frac{3}{4} \, \log\left(4\right) + \frac{1}{2} \, \log\left(2\right) + \frac{3}{4} \, \log\left(\left(8 \cdot 2^{\frac{1}{3}}\right) - 4 \cdot 2^{\frac{2}{3}} + 4\right) + \frac{1}{4} \, \log\left(2^{\frac{2}{3}} - 2 \cdot 2^{\frac{1}{3}} + 4\right) + \log\left(2^{\frac{1}{3}} + 2\right)$

-1.428734239102843727769729435381676199314887493528498326518907304034540\
794883422744594310027325215037e-101

-1.428734239102843727769729435381676199314887493528498326518907304034540794883422744594310027325215037e-101

# In polar coordinates... # Under z = e^(-cx^2-cy^2) # inside d^2 < x^2+y^2 < b*a^2 var('r,theta,a,b,c,d') assume(c>1,b>1) I1 = integrate(e^(-c*r^2)*r,r,d,sqrt(b)*a) I = integrate(I1,theta,0,2*pi) lim(I,a=oo)

pi*e^(-c*d^2)/c

pi*e^(-c*d^2)/c

222 - Topic 14.5 - Switching order of integration

1707 days ago by Professor222

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