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Question Number 220972 Answers: 1 Comments: 2

∫_(−∞) ^(+∞) ∫_(−∞) ^(+∞) (1/((x^2 +y^2 +4)^2 )) dxdy x=rcos(θ) y=rsin(θ) ∣∣J∣∣=rdrdθ ∫_0 ^( 2π) ∫_0 ^( ∞) (r/((r^2 +4)^2 ))drdθ=(1/8)∫_0 ^( 2π) dθ=(π/4) Q. if ∫_0 ^( ∞) ∫_0 ^( ∞) (1/((x^2 +y^2 +4)^2 )) dxdy 0≤r<∞ , 0≤θ≤(π/2).....??? I really confuse how could i select interval of integral

$$\int_{−\infty} ^{+\infty} \int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$${x}={r}\mathrm{cos}\left(\theta\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right) \\ $$$$\mid\mid\boldsymbol{{J}}\mid\mid={r}\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{r}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\mathrm{d}{r}\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\mathrm{d}\theta=\frac{\pi}{\mathrm{4}} \\ $$$${Q}.\:\mathrm{if}\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\mathrm{0}\leq{r}<\infty\:,\:\mathrm{0}\leq\theta\leq\frac{\pi}{\mathrm{2}}.....??? \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{confuse}\:\mathrm{how}\:\mathrm{could}\:\:\mathrm{i}\:\mathrm{select}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{integral} \\ $$

Question Number 220971 Answers: 1 Comments: 6

Question Number 221680 Answers: 2 Comments: 4

I suspect π=i∫_i ^(−1) (((z−1)dz)/( z(√(z^2 +1)))) someone please help confirm or reject!

$${I}\:{suspect} \\ $$$$\pi={i}\underset{\boldsymbol{{i}}} {\overset{−\mathrm{1}} {\int}}\frac{\left({z}−\mathrm{1}\right){dz}}{\:{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${someone}\:{please}\:{help}\:{confirm}\:{or}\:{reject}! \\ $$

Question Number 220968 Answers: 0 Comments: 0

Question Number 220964 Answers: 0 Comments: 0

Find the general solution of the differential equation x^2 (d^3 y/dx^3 ) + x(d^2 y/dx^2 )−6(dy/dx)+6(y/x)=((x ln x+1)/x^2 ),[x>0]

$${Find}\:{the}\:{general}\:{solution}\:{of}\:{the}\:{differential}\:{equation} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{6}\frac{{dy}}{{dx}}+\mathrm{6}\frac{{y}}{{x}}=\frac{{x}\:\mathrm{ln}\:{x}+\mathrm{1}}{{x}^{\mathrm{2}} },\left[{x}>\mathrm{0}\right] \\ $$

Question Number 220963 Answers: 1 Comments: 0

Let f:R^2 →R be defined by f(x,y)={(y/(sin y_( 1, y=0) )), y≠0 Then the integral (1/π^2 )∫_(x=0) ^1 ∫_(y=sin^(−1) x) ^(π/2) f(x,y)dy dx correct upto three decimal places,is...

$${Let}\:{f}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}\:{be}\:{defined}\:{by}\:{f}\left({x},{y}\right)=\left\{\frac{{y}}{\underset{\:\:\mathrm{1},\:{y}=\mathrm{0}} {\mathrm{sin}\:{y}}},\:{y}\neq\mathrm{0}\right. \\ $$$${Then}\:{the}\:{integral}\:\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\underset{{x}=\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{{y}=\mathrm{sin}^{−\mathrm{1}} {x}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{f}\left({x},{y}\right){dy}\:{dx}\:{correct}\:{upto}\:{three}\:{decimal}\:{places},{is}... \\ $$

Question Number 220958 Answers: 1 Comments: 1

for x, y, z >0 find the maximum of x^m y^n z^k subject to ax+by+cz=d.

$${for}\:{x},\:{y},\:{z}\:>\mathrm{0}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${x}^{{m}} {y}^{{n}} {z}^{{k}} \:{subject}\:{to}\:{ax}+{by}+{cz}={d}. \\ $$

Question Number 220950 Answers: 1 Comments: 0

∫_( 0) ^( π) ∫_( 0) ^( 1) ∫_( 0) ^( π) sin^( 2) x + y sin z dxdydz = (1/2) π (2 + π)

$$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\pi} \int_{\:\mathrm{0}} ^{\:\mathrm{1}} \int_{\:\mathrm{0}} ^{\:\:\pi} \:\mathrm{sin}^{\:\mathrm{2}} \:{x}\:+\:{y}\:\mathrm{sin}\:{z}\:{dxdydz}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\pi\:\left(\mathrm{2}\:+\:\pi\right)\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220948 Answers: 1 Comments: 0

∫ x^2 (√(5−x^6 ))dx

$$\int\:{x}^{\mathrm{2}} \sqrt{\mathrm{5}−{x}^{\mathrm{6}} }{dx} \\ $$

Question Number 220947 Answers: 1 Comments: 0

Σ_(k=1) ^(13) (1/(sin ((π/4)+(((k−1)π)/6))sin ((π/4)+((kπ)/6))))