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# MTH 101 Assignment # 2 Question No 1 Solution

Please correct me if i am wrong

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### Replies to This Discussion

plzzz 1 day left, send full solution of mth-101 assignment-2

plzz koi full solution upload kro

aray bhai koi to kuch clear cut solution uplod kro yar suggestions ni chaiye han

proper solution bhejo dosto

kia koi bi maths ka expert nai idar??????????

Hey folks,

[f(x) - g(x)]dx

[x^2 - 3x - (-2)]

[x^2 - 3x + 2]

Please note the sign of + before 2.

ali bhai ap ne 3rd last step mein 3/4 ko 4/3 ksey kia hai aur  27/64 q likha hai plz explain

x* ka kia mtlb ha i mean ye kun use kia gya ha

plzzz 1 day left, send full solution of mth-101 assignment-2

ye samajhne ka waqt nahi k kia kyun kese hua bs kuch rad o badal se assignment likh k bhejo sabhi

and to be honest i think koi bi confident nahi hai is sab solutions k baarey mein ajeeb bat hai koi bi mathematician nahi pooray group mein jo certainty se verify ya modify kr sake in suggestions ko so just contented rahein jo hai idar

ali raza marshal bhai agr ap mjhe thoda sa guide kr detey in points k barey main tu ju main ne pehly pochey tu apki bari inayat hoti ..... khair koe baat  nahi actually kafi bizi honey ki waja se solution mein nahi de ska.

kon sa points

Question # 3:                                                                            Marks 5

Find the volume of the spheroid formed by the revolution of the area bounded by the ellipse  about the major axis.

By using parametric form…..

x = aCos t
y = bSin t,                    0 ≤ t ≤ π

Which gives the half ellipse with positive y

dx/dt = - aSint
dy/dt = bCos t

V = π∫(aCos t)²(bCos t) dt            on    [0,π]

Thanks to the symmetry w.r.t  y-axis we have

V = 2πa²b∫(Cos t)²(Cos t) dt         on    [0,π/2]

= (*)

∫(cos t)³ dt = ∫(cos t)(1 - (sin t)²) dt

Substitute
Sin t = u
Cos t dt = du

when t = 0,          u = 0
when t = π/2,       u = 1

(*)
= 2πa²b ∫(1 - u²) du                         on     [0,1]

A primitive is:

F (u) = 2πa²b(u - (1/3) u³)

Then volume is

V = F (1) - F (0) = (4/3)πa²b

Hence proved.

The volume of a spheroid (of any kind) is $\frac{4}{3}\pi a^2b \approx 4.19\, a^2b$

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