Ankit Pandey ♦♦ 10
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#### Posts by Ankit Pandey

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... Solution: Given curve: $r^{2}=a^{2} \cos 2 \theta \quad$ is lemniscate. The density at any point is proportional to the square of dist. From the pole. Distance from the pole =r $\quad \therefore$ Density $\propto r^{2}$ $\therefore$ Density $=k . r^{2}$ The mass of the lemniscate is given by, ...
written 3 days ago by Ankit Pandey ♦♦ 10
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... Solution: \begin{aligned} \text { Let } 1 &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d x \ d y \ d z \\ &=\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} \frac{1}{(x+y+z+1)^{3}} d z \ d y \ d x \end{aligned} \begin{aligned} &=\int_{0}^{1} \int_{0}^{1-x}\left ... written 3 days ago by Ankit Pandey ♦♦ 10 1 answer 24 views 1 answers ... Solution: LetI=\int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{x d y d x}{\sqrt{x^{2}+y^{2}}}$Region of integration is:$\quad x \leq y \leq \sqrt{2-x^{2}}\\ 0 \leq x \leq 1$Curves:$(i) y=x$line$\quad(i i) x=0, x=1 \quad$lines parallel to the **y** axis (iii)$y=\sqrt{2-x^{2}} \Rightarrow x^{ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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... Solution: Let $I(\mathrm{a})=\int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x---(1)$ Taking 'a' as parameter diff. w.r.t. a, $\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{\infty} e^{-\left(x^{2}+\frac{a^{2}}{x^{2}}\right)} d x$ Apply D.U.I.S rule, $\frac{d I(a)}{d a}=\int_{0}^{\i ... written 3 days ago by Ankit Pandey ♦♦ 10 1 answer 23 views 1 answers ... Solution: Given curve: Cycloid$x=a(\theta+\sin \theta), y=a(1-\cos \theta)$![enter image description here][1] The length of given curve is:$\mathrm{S}=\int_{\theta_{1}}^{\theta_{2}} \sqrt{\left(\frac{d x}{d \theta}\right)^{2}+\left(\frac{d y}{d \theta}\right)^{2}} \mathrm{d} \theta\begin{ ...
written 3 days ago by Ankit Pandey ♦♦ 10
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... Solution: let $\quad I=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x$ Taking 'a' as parameter, $\quad I(a)=\int_{0}^{1} \frac{x^{a}-1}{\log x} d x---(1)$ differentiate w.r.t **a** $\frac{d I(a)}{d a}=\frac{d}{d a} \int_{0}^{1} \frac{x^{a}-1}{\log x} d x$ $\therefore \frac{d I(a)}{d a}=\int_{0}^{1} \ ... written 3 days ago by Ankit Pandey ♦♦ 10 1 answer 30 views 1 answers ... Solution:$\begin{aligned} \text { let } \mathrm{I} &=\int_{0}^{1} \int_{0}^{x^{2}} e^{\frac{y}{x}} d y d x \\ &=\int_{0}^{1}\left[\frac{e^{y}}{\frac{1}{x}}\right]_{0}^{x^{2}} d x \end{aligned}=\int_{0}^{1} \frac{\left(e^{x}-1\right)}{\frac{1}{x}} d x=\int_{0}^{1} x \cdot e^{x} d x- ...
written 3 days ago by Ankit Pandey ♦♦ 10
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... Solution: $\left(D^{2}-6 D+9\right) y=\frac{e^{3 x}}{x^{2}}$ For complementary solution, $f(D)=0$ $\therefore\left(D^{2}-6 D+9\right)=0$ Roots are: $D=3,3 \quad$ Real roots but repeatative. The complementary solution of given diff. eqn is, $\quad \therefore y_{c}=\left(c_{1}+x c_{2}\right) e^ ... written 3 days ago by Ankit Pandey ♦♦ 10 1 answer 22 views 1 answers ... Solution: Let$\mathrm{V}=\iiint x^{2} d x \ d y\ d z$Region of integration is volume bounded by the planes$x=0, y=0, z=0$And$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$Put$x=a u, y=b v, z=c w\therefore \quad d x \ d y \ d z=a b c$du.dv.dw ![enter image description here][1] The inters ... written 3 days ago by Ankit Pandey ♦♦ 10 1 answer 16 views 1 answers ... Solution: Let$I=\int_{0}^{2} \int_{\sqrt{2 y}}^{2} \frac{x^{2}}{\sqrt{x^{4}-4 y^{2}}} d x \ d y\begin{aligned} \text { Region of integration : } & \sqrt{2 y} \leq x \leq 2 \\ 0 & \leq y \leq 2 \end{aligned}\begin{aligned} \text { Curves: } & \text { (i) } x=2, y=2, y=0 \text { ...