Evaluate the given integral by changing to polar coordinates.64-x2-y2 dARwhere R ={(x, y) | x2 + y2 %3C= 64, x %3C= 0}. | Homework.Study.com
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SOLVED:Find (4 cis (π)/(6))^3. a. 12 cis (π)/(2) b. 64 cis (π^3)/(216) c. 64 cis (π)/(2) d. 12 cis (π)/(18)
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For a positive integer n , let fn(theta)=(tantheta//2)(1+sectheta)(1+sec2theta)(1+sec4theta).......... (1+sec2^ntheta). Then (a)f2(pi/(16))=1 (b) f3(pi/(32))=1 (c)f4(pi/(64))=1 (d) f5(pi/(128))=1
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![tan^-1 [ cos x1 + sin x ] is equal to](https://dwes9vv9u0550.cloudfront.net/images/3616471/165cccf1-7df9-468e-aa5d-ceeb0aec94a7.jpg "tan^-1 [ cos x1 + sin x ] is equal to")
tan^-1 [ cos x1 + sin x ] is equal to
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A lamina of mass $M$ in the shape of a quarter of the disc of radius $R$ as shown in figure. The moment of inertia of this lamina about axis $AB$ is (
![a) = −i dt (d − 3) 2 Γ 2 [ d−3 2 ] 128(d − 2) 2 π d−1 | Download Scientific Diagram](https://www.researchgate.net/publication/23503561/figure/fig1/AS:588227446603779@1517255760835/Fig3a-i-dt-d-3-2-G-2-d-3-2-128d-2-2-p-d-1.png "a) = −i dt (d − 3) 2 Γ 2 [ d−3 2 ] 128(d − 2) 2 π d−1 | Download Scientific Diagram")
a) = −i dt (d − 3) 2 Γ 2 [ d−3 2 ] 128(d − 2) 2 π d−1 | Download Scientific Diagram
1000 small water drops each of radius 'r' and charge 'q' coalesce together to one spherical drop. The potential of the big drop is larger than that of the smaller drops by
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