I-0-L-2-Rz-2-d-2-z-2-d-2-z-2-R-2-dz-Find-I-

Question Number 47737 by ajfour last updated on 14/Nov/18

I = \int_{0}^{L / 2} \frac{{R z}^{2}}{(d^{2} + z^{2}) \sqrt{d^{2} + z^{2} - R^{2}}} d z

F i n d I .

Answered by MJS last updated on 14/Nov/18

R \int \frac{z^{2}}{(z^{2} + d^{2}) \sqrt{z^{2} + d^{2} - R^{2}}} d z =

= R \int \frac{d z}{\sqrt{z^{2} + d^{2} - R^{2}}} - d^{2} R \int \frac{d z}{(z^{2} + d^{2}) \sqrt{z^{2} + d^{2} - R^{2}}}

R \int \frac{d z}{\sqrt{z^{2} + d^{2} - R^{2}}} =

[t = \frac{z}{\sqrt{d^{2} - R^{2}}} \to d z = \sqrt{d^{2} - R^{2}} d t]

= R \int \frac{d t}{\sqrt{t^{2} + 1}} = R \ln (t + \sqrt{t^{2} + 1}) = R \ln \frac{z + \sqrt{z^{2} + d^{2} - R^{2}}}{\sqrt{d^{2} - R^{2}}}

- d^{2} R \int \frac{d z}{(z^{2} + d^{2}) \sqrt{z^{2} + d^{2} - R^{2}}} =

[z = \sqrt{d^{2} - R^{2}} \tan u \to u = \arctan \frac{z}{\sqrt{d^{2} - R^{2}}}; d z = \sqrt{d^{2} - R^{2}} \sec^{2} u d u]

= - d^{2} R \int \frac{\sqrt{d^{2} - R^{2}} \sec^{2} u}{(d^{2} + (d^{2} - R^{2}) \tan^{2} u) \sqrt{d^{2} - R^{2} + (d^{2} - R^{2}) \tan^{2} u}} d u =

= - d^{2} R \int \frac{\sec u}{d^{2} + (d^{2} - R^{2}) \tan^{2} u} d u = - d^{2} R \int \frac{\cos u}{d^{2} - R^{2} \sin^{2} u} d u =

[v = \sin u \to d u = \frac{d v}{\cos u}]

= - d^{2} R \int \frac{d v}{d^{2} - R^{2} v^{2}} = - d^{2} R \int \frac{d v}{(d + R v) (d - R v)} =

= \frac{d R}{2} \int \frac{d v}{d - R v} - \frac{d R}{2} \int \frac{d v}{d + R v} =

= \frac{d}{2} \ln \frac{d - R v}{d + R v} = \frac{d}{2} \ln \frac{d - R \sin u}{d + R \sin u} =

= \frac{d}{2} \ln \frac{d - R \sin \arctan \frac{z}{\sqrt{d^{2} - R^{2}}}}{d + R \sin \arctan \frac{z}{\sqrt{d^{2} - R^{2}}}} =

= \frac{d}{2} \ln \frac{d - R \frac{z}{\sqrt{z^{2} + d^{2} - R^{2}}}}{d + R \frac{z}{\sqrt{z^{2} + d^{2} - R^{2}}}} = \frac{d}{2} \ln \frac{d \sqrt{z^{2} + d^{2} - R^{2}} - R z}{d \sqrt{z^{2} + d^{2} - R^{2}} + R z}

\int \frac{{R z}^{2}}{(z^{2} + d^{2}) \sqrt{z^{2} + d^{2} - R^{2}}} d z =

R \ln ∣ \frac{z + \sqrt{z^{2} + d^{2} - R^{2}}}{\sqrt{d^{2} - R^{2}}} ∣ + \frac{d}{2} \ln ∣ \frac{R z - d \sqrt{z^{2} + d^{2} - R^{2}}}{R z + d \sqrt{z^{2} + d^{2} - R^{2}}} ∣ + C

Commented by MJS last updated on 14/Nov/18

it only works for d > R

Commented by MJS last updated on 14/Nov/18

I have no time to check it for typos and other

minor mistakes but this is the path to solve it

Commented by ajfour last updated on 14/Nov/18

T h a n k y o u S i r, u n d e r s t o o d t h e

m e t h o d, q u i t e n a t u r a l .

Commented by ajfour last updated on 14/Nov/18

o f c o u r s e, i t s h o u l d b e s o, p l e a s e

r e s o l v e o u r i s s u e; i f a t a l l t h i s

i n t e g r a l a r i s e s o r n o t ?

P l e a s e v i e w Q . 47728

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