Need Help !!

Find the volume of the solid generated by revolving cissoid y2=x32a−x{ y }^{ 2 }=\frac { { x }^{ 3 } }{ { 2 }{ a }-{ x } } about its asymptote x = 2a

I tried this problem many times but still not able to get it

#Calculus

Note by A Former Brilliant Member
3 years, 2 months ago

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Comments

Reflect the cissoid, and consider the volume of revolution of y2  =  (2−x)3x0≤x≤2 y^2 \; = \; \frac{(2-x)^3}{x} \hspace{2cm} 0 \le x \le 2 about the yy-axis. This is the case a=1a=1. Then 2y dy  =  −2(1+x)(2−x)2x2 dx 2y\,dy \; = \; -\frac{2(1+x)(2-x)^2}{x^2}\,dx so that dy  =  −(1+x)(2−x)12x32 dx dy \; = \; -\frac{(1+x) (2-x)^{\frac12}}{x^{\frac32}}\,dx and hence the volume of revolution is V=  π∫0∞πx2 dy  =  π∫02(1+x)x(2−x) dx  =  π∫−11(x+2)1−x2 dx=  2π∫−111−x2 dx  =  4π∫011−x2 dx  =  4π∫012πcos⁡2θ dθ=  2π[12sin⁡2θ+θ]012π  =  π2 \begin{aligned} V & = \; \pi \int_0^\infty \pi x^2\,dy \; = \; \pi \int_0^{2} (1+x)\sqrt{x(2-x)}\,dx \; = \; \pi\int_{-1}^1 (x+2)\sqrt{1-x^2}\,dx \\ & = \; 2\pi\int_{-1}^1 \sqrt{1-x^2}\,dx \; = \; 4\pi\int_0^1 \sqrt{1-x^2}\,dx \; = \; 4\pi\int_0^{\frac12\pi} \cos^2\theta\,d\theta \\ & = \; 2\pi \Big[\tfrac12\sin2\theta + \theta\Big]_0^{\frac12\pi} \; =\; \pi^2 \end{aligned} Scaling by aa gives the general volume of π2a3\pi^2a^3.

Mark Hennings - 3 years, 1 month ago

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@Mark Hennings, thank u sir got my mistake

A Former Brilliant Member - 3 years, 1 month ago

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@Mark Hennings Sir please help me with the integral given below \large\int_0^\infty\dfrac{x^3\ln(x)}{(1+x^4)^3

A Former Brilliant Member - 3 years ago

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@A Former Brilliant Member – Put y=x4y = x^4. Then I  =  ∫0∞x3ln⁡x(1+x4)3 dx  =  116∫0∞ln⁡y(1+y)3 dy  =  116dda∫0∞ya−1(1+y)3 dy∣a=1I \; = \; \int_0^\infty \frac{x^3 \ln x}{(1+x^4)^3}\,dx \; = \; \tfrac{1}{16}\int_0^\infty \frac{\ln y}{(1+y)^3}\,dy \; = \; \tfrac{1}{16} \frac{d}{da} \int_0^\infty \frac{y^{a-1}}{(1+y)^3}\,dy \Big|_{a=1} and hence I=  116ddaB(a,3−a)∣a=1  =  132dda(Γ(a)Γ(3−a))∣a=1=  132Γ(a)Γ(3−a)(ψ(a)−ψ(3−a))∣a=1  =  132(ψ(1)−ψ(2))  =  −132 \begin{aligned} I & = \; \tfrac{1}{16}\frac{d}{da}B(a,3-a) \Big|_{a=1} \; = \; \tfrac{1}{32}\frac{d}{da}\Big(\Gamma(a)\Gamma(3-a)\Big)\Big|_{a=1} \\ & = \; \frac{1}{32} \Gamma(a)\Gamma(3-a)\big(\psi(a) - \psi(3-a)\big)\Big|_{a=1} \; =\; \tfrac{1}{32}(\psi(1) - \psi(2)) \; = \; -\tfrac{1}{32} \end{aligned}

Mark Hennings - 3 years ago

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