Question Number 101293 by mhmd last updated on 01/Jul/20

Commented bymhmd last updated on 01/Jul/20

help me sir

Commented bysmridha last updated on 01/Jul/20

unit vector along A^→  isA^� =(1/(√(21)))(4i^� −2j^� −k^� )  gradient of scaler field at(−1,1,1)  [▽∅]_((−1,1,1)) =[(8x+z)i^� −z^2 j^� +(x−2zy)k^� ]_((−1,11))                        =[−7i^� −j^� −3k^� ]  so the directional derivative is  =[▽∅]_((−1,1,1)) .A^� =(1/(√(21)))(−28+2+3)  =((−23)/(√(21))).

Commented bymhmd last updated on 01/Jul/20

thank you sir .very very thank

Commented bysmridha last updated on 01/Jul/20

∫_0 ^2 dx[ysin(x)+sin(y)]_0 ^𝛑   =𝛑∫_0 ^2 sin(x)dx=−𝛑[cos(x)]_0 ^2 =𝛑(1−cos2)  or 2𝛑sin^2 (1)

Commented bysmridha last updated on 01/Jul/20

welcome...

Commented bymhmd last updated on 01/Jul/20

thank you sir

Commented bysmridha last updated on 01/Jul/20

dy−32x^2 sin(2x)dx=0  integrating both sides ...  y−32[x^2 ∫sin2x+∫x.cos(2x)]=C  y−32[−((x^2 cos2x)/2)+((xsin(2x))/2)+(1/4)cos2x]=C  y+16x^2 cos2x−16xsin2x−8cos2x=C

Commented bymhmd last updated on 01/Jul/20

sir can you help me in question 4 pleas

Answered by mr W last updated on 01/Jul/20

Q4  see Q88758    Method I  4x^2 =8−4x^2   8x^2 −8=0  a=8, b=0, c=−8  Δ=b^2 −4ac=0^2 −4×8×(−8)=256  bounded area=((Δ(√Δ))/(6a^2 ))=((256(√(256)))/(6×8^2 ))=((32)/3)    METHOD II  bounded area=2×(2/3)×2×4=((32)/3)