(i)
∫02∫x2x(x4+y2)dydx=∫02[x4y+3y3]2xxdx==∫02(x4(2x)+3(2x)3−x4(x)−3x3)dx==∫02(x5+37x3)dx=[6x6+127x4]20==626+127(2)4−0=20 ∫02∫x2x(x4+y2)dydx=20
(ii) z=25−x2−y2
The region E in the xy-plane is the disk 0≤x2+y2≤25 (disk or radius 5 centered at the origin).
For this problem, z=f(x,y)=25−x2−y2.
fx=−2x,fy=−2y
1+(fx)2+(fy)2=1+(−2x)2+(−2y)2=1+4x2+4y2
The region E={(x,y)∣x2+y2≤25} is {(r,θ∣0≤r≤5,0≤θ≤2π} in polar coordinates.
1+4x2+4y2=1+4r2
Hence, the surface area A is given by
A=∫∫E1+(fx)2+(fy)2dxdy==∫02π∫051+4r2rdrdθ==∫02π[12(1+4r2)3/2]50dθ==122π((1+4(5)2)3/2−1)=6π((101)3/2−1) (iii)
f(x,y)=x2+y2−6xy+6x+3y−4
Partial derivatives
fx=2x−6y+6
fy=2y−6x+3
To find the critical points, we solve
fx=0=>2x−6y+6=0
fy=0=>2y−6x+3=0
x=−16111,y=−1621
We find the critical point (−16111,−1621)
To find the nature of the critical point, we apply the second derivative test. We have
A=fxx=2,B=fxy=−6,C=fyy=2
At the point (−16111,−1621), we have
D=AC−B2=2(2)−(−6)2=−32<0 Then the point (−16111,−1621) is a saddle point.
(iv)Check whether the following functions are homogeneous or not.
(a)
f(x,y)=yx+2x3y+sinyxf(tx,ty)=tytx+2tx3ty+sintytx==yx+2x3y+sinyx=t0f(x,y) A function f(x,y) is homogeneous.
(b)
f(x,y)=x4+4x2+x2y2f(tx,ty)=(tx)4+4(tx)2+(tx)2(ty)2=t4x4+4t2x2+x2y2
A function f(x,y) is not homogeneous.
(v)
f(x,y)=x5+10x3y3+8y4Evaluate fxy at a point (x,y)
fx(x,y)=5x4+30x2y3fxy(x,y)=90x2y2 Verify that the function f satisfies the requirements of Schwarz’s theorem and hence evaluate f(x,y)
Since 90x2y2 is a polynomial, fxy(x,y) is a continuous function.
Further, fy(x,y)=30x3y2+32y3 exists. Hence f satisfies the conditions of Schwarz's theorem and so
fyx(x,y)=fxy(x,y)=90x2y2
Comments