Question #104428
(i)Find the integral of f(x,y ) x^4+y^2 over the region bounded by y=x , y= 2x and x=2
(ii)Find the surface area of the portion of the paraboloid z=25–x^2–y^2 which lies
above the xy -plane.
(iii)Locate and classify the stationary points of the function
f( x,y ) x^2+y^2–6xy +6x +3y –4
(iv)Check whether the following functions are homogeneous or not.
(a)x/y +3y/2x +sin√(x/y)
(b)x^4 +4x^2 +y^2)x^2

(v)Evaluate f(xy )at a point (x,y)for the function f defined by
f(x,y)=x^5 +10x^3 y^3 +8y^4
Verify that the function f satisfies the requirements of Schwarz’s theorem and
hence evaluate f(x,y)
1
Expert's answer
2020-03-06T10:11:25-0500

(i)


02x2x(x4+y2)dydx=02[x4y+y33]2xxdx=\displaystyle\int_{0}^2\displaystyle\int_{x}^{2x}(x^4+y^2)dydx=\displaystyle\int_{0}^2\bigg[x^4y+{y^3 \over 3}\bigg]\begin{matrix} 2x \\ x \end{matrix}dx==02(x4(2x)+(2x)33x4(x)x33)dx==\displaystyle\int_{0}^2\bigg(x^4(2x)+{(2x)^3 \over 3}-x^4(x)-{x^3 \over 3}\bigg)dx==02(x5+7x33)dx=[x66+7x412]20==\displaystyle\int_{0}^2\big(x^5+{7x^3 \over 3}\big)dx=\big[{x^6 \over6}+{7x^4 \over 12}\big]\begin{matrix} 2 \\ 0 \end{matrix}==266+7(2)4120=20={2^6 \over 6}+{7(2)^4 \over 12}-0=20

(ii) z=25x2y2z=25-x^2-y^2

The region EE in the xy-plane is the disk 0x2+y2250\leq x^2+y^2\leq25 (disk or radius 5 centered at the origin).

For this problem, z=f(x,y)=25x2y2.z=f(x, y)=25-x^2-y^2.

fx=2x,fy=2yf_x=-2x, f_y=-2y

1+(fx)2+(fy)2=1+(2x)2+(2y)2=1+4x2+4y2\sqrt{1+(f_x)^2+(f_y)^2}=\sqrt{1+(-2x)^2+(-2y)^2}=\sqrt{1+4x^2+4y^2}

The region E={(x,y)x2+y225}E=\{(x,y)|x^2+y^2\leq25\} is {(r,θ0r5,0θ2π}\{(r,\theta|0\leq r\leq5, 0\leq \theta\leq2\pi\} in polar coordinates.

1+4x2+4y2=1+4r2\sqrt{1+4x^2+4y^2}=\sqrt{1+4r^2}

Hence, the surface area AA is given by


A=E1+(fx)2+(fy)2dxdy=A=\int\int_{E}\sqrt{1+(f_x)^2+(f_y)^2}dxdy==02π051+4r2rdrdθ==\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{5}\sqrt{1+4r^2}rdrd\theta==02π[(1+4r2)3/212]50dθ==\displaystyle\int_{0}^{2\pi}\bigg[{(1+4r^2)^{3/2} \over 12}\bigg]\begin{matrix} 5 \\ 0 \end{matrix}d\theta==2π12((1+4(5)2)3/21)=π6((101)3/21)={2\pi \over 12}\big((1+4(5)^2)^{3/2}-1\big)={\pi \over 6}\big((101)^{3/2}-1\big)

(iii)

f(x,y)=x2+y26xy+6x+3y4f(x,y)=x^2+y^2-6xy+6x+3y-4

Partial derivatives

fx=2x6y+6f_x=2x-6y+6

fy=2y6x+3f_y=2y-6x+3

To find the critical points, we solve

fx=0=>2x6y+6=0f_x=0=>2x-6y+6=0

fy=0=>2y6x+3=0f_y=0=>2y-6x+3=0


x=11116,y=2116x=-{111 \over 16}, y=-{21 \over 16}

We find the critical point (11116,2116)(-{111 \over 16},-{21 \over 16})

To find the nature of the critical point, we apply the second derivative test. We have

A=fxx=2,B=fxy=6,C=fyy=2A=f_{xx}=2, B=f_{xy}=-6, C=f_{yy}=2

At the point (11116,2116),(-{111 \over 16},-{21 \over 16}), we have


D=ACB2=2(2)(6)2=32<0D=AC-B^2 =2(2)-(-6)^2=-32<0

Then the point (11116,2116)(-{111 \over 16},-{21 \over 16}) is a saddle point.


(iv)Check whether the following functions are homogeneous or not.

(a)


f(x,y)=xy+3y2x+sinxyf(x, y)={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}}f(tx,ty)=txty+3ty2tx+sintxty=f(tx, ty)={tx \over ty}+{3ty \over 2tx}+\sin{\sqrt{{tx \over ty}}}==xy+3y2x+sinxy=t0f(x,y)={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}}=t^0 f(x, y)

A function f(x,y)f(x,y) is homogeneous.


(b)


f(x,y)=x4+4x2+y2x2f(x, y)=x^4+4x^2+{y^2 \over x^2}f(tx,ty)=(tx)4+4(tx)2+(ty)2(tx)2=t4x4+4t2x2+y2x2f(tx, ty)=(tx)^4+4(tx)^2+{(ty)^2 \over (tx)^2}=t^4x^4+4t^2x^2+{y^2 \over x^2}

A function f(x,y)f(x,y) is not homogeneous.


(v)


f(x,y)=x5+10x3y3+8y4f(x,y)=x^5+10x^3y^3+8y^4

Evaluate fxyf_{xy} at a point (x,y)(x,y)


fx(x,y)=5x4+30x2y3f_x(x,y)=5x^4+30x^2y^3fxy(x,y)=90x2y2f_{xy}(x,y)=90x^2y^2

Verify that the function ff satisfies the requirements of Schwarz’s theorem and hence evaluate fyxf_{yx}

Since 90x2y290x^2y^2 is a polynomial, fxy(x,y)f_{xy}(x,y) is a continuous function. 

Further, fy=30x3y2+32y3f_y=30x^3y^2+32y^3 exists. Hence ff satisfies the conditions of Schwarz's theorem and so 


fyx(x,y)=fxy(x,y)=90x2y2f_{yx}(x,y)=f_{xy}(x,y)=90x^2y^2

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