(i)
∫ 0 2 ∫ x 2 x ( x 4 + y 2 ) d y d x = ∫ 0 2 [ x 4 y + y 3 3 ] 2 x x d x = \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= ∫ 0 2 ∫ x 2 x ( x 4 + y 2 ) d y d x = ∫ 0 2 [ x 4 y + 3 y 3 ] 2 x x d x = = ∫ 0 2 ( x 4 ( 2 x ) + ( 2 x ) 3 3 − x 4 ( x ) − x 3 3 ) d x = =\displaystyle\int_{0}^2\bigg(x^4(2x)+{(2x)^3 \over 3}-x^4(x)-{x^3 \over 3}\bigg)dx= = ∫ 0 2 ( x 4 ( 2 x ) + 3 ( 2 x ) 3 − x 4 ( x ) − 3 x 3 ) d x = = ∫ 0 2 ( x 5 + 7 x 3 3 ) d x = [ x 6 6 + 7 x 4 12 ] 2 0 = =\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}= = ∫ 0 2 ( x 5 + 3 7 x 3 ) d x = [ 6 x 6 + 12 7 x 4 ] 2 0 = = 2 6 6 + 7 ( 2 ) 4 12 − 0 = 20 ={2^6 \over 6}+{7(2)^4 \over 12}-0=20 = 6 2 6 + 12 7 ( 2 ) 4 − 0 = 20 (ii) z = 25 − x 2 − y 2 z=25-x^2-y^2 z = 25 − x 2 − y 2
The region E E E in the xy-plane is the disk 0 ≤ x 2 + y 2 ≤ 25 0\leq x^2+y^2\leq25 0 ≤ x 2 + y 2 ≤ 25 (disk or radius 5 centered at the origin).
For this problem, z = f ( x , y ) = 25 − x 2 − y 2 . z=f(x, y)=25-x^2-y^2. z = f ( x , y ) = 25 − x 2 − y 2 .
f x = − 2 x , f y = − 2 y f_x=-2x, f_y=-2y f x = − 2 x , f y = − 2 y
1 + ( f x ) 2 + ( f y ) 2 = 1 + ( − 2 x ) 2 + ( − 2 y ) 2 = 1 + 4 x 2 + 4 y 2 \sqrt{1+(f_x)^2+(f_y)^2}=\sqrt{1+(-2x)^2+(-2y)^2}=\sqrt{1+4x^2+4y^2} 1 + ( f x ) 2 + ( f y ) 2 = 1 + ( − 2 x ) 2 + ( − 2 y ) 2 = 1 + 4 x 2 + 4 y 2
The region E = { ( x , y ) ∣ x 2 + y 2 ≤ 25 } E=\{(x,y)|x^2+y^2\leq25\} E = {( x , y ) ∣ x 2 + y 2 ≤ 25 } is { ( r , θ ∣ 0 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π } \{(r,\theta|0\leq r\leq5, 0\leq \theta\leq2\pi\} {( r , θ ∣0 ≤ r ≤ 5 , 0 ≤ θ ≤ 2 π } in polar coordinates.
1 + 4 x 2 + 4 y 2 = 1 + 4 r 2 \sqrt{1+4x^2+4y^2}=\sqrt{1+4r^2} 1 + 4 x 2 + 4 y 2 = 1 + 4 r 2
Hence, the surface area A A A is given by
A = ∫ ∫ E 1 + ( f x ) 2 + ( f y ) 2 d x d y = A=\int\int_{E}\sqrt{1+(f_x)^2+(f_y)^2}dxdy= A = ∫ ∫ E 1 + ( f x ) 2 + ( f y ) 2 d x d y = = ∫ 0 2 π ∫ 0 5 1 + 4 r 2 r d r d θ = =\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{5}\sqrt{1+4r^2}rdrd\theta= = ∫ 0 2 π ∫ 0 5 1 + 4 r 2 r d r d θ = = ∫ 0 2 π [ ( 1 + 4 r 2 ) 3 / 2 12 ] 5 0 d θ = =\displaystyle\int_{0}^{2\pi}\bigg[{(1+4r^2)^{3/2} \over 12}\bigg]\begin{matrix}
5 \\
0
\end{matrix}d\theta= = ∫ 0 2 π [ 12 ( 1 + 4 r 2 ) 3/2 ] 5 0 d θ = = 2 π 12 ( ( 1 + 4 ( 5 ) 2 ) 3 / 2 − 1 ) = π 6 ( ( 101 ) 3 / 2 − 1 ) ={2\pi \over 12}\big((1+4(5)^2)^{3/2}-1\big)={\pi \over 6}\big((101)^{3/2}-1\big) = 12 2 π ( ( 1 + 4 ( 5 ) 2 ) 3/2 − 1 ) = 6 π ( ( 101 ) 3/2 − 1 ) (iii)
f ( x , y ) = x 2 + y 2 − 6 x y + 6 x + 3 y − 4 f(x,y)=x^2+y^2-6xy+6x+3y-4 f ( x , y ) = x 2 + y 2 − 6 x y + 6 x + 3 y − 4
Partial derivatives
f x = 2 x − 6 y + 6 f_x=2x-6y+6 f x = 2 x − 6 y + 6
f y = 2 y − 6 x + 3 f_y=2y-6x+3 f y = 2 y − 6 x + 3
To find the critical points, we solve
f x = 0 = > 2 x − 6 y + 6 = 0 f_x=0=>2x-6y+6=0 f x = 0 => 2 x − 6 y + 6 = 0
f y = 0 = > 2 y − 6 x + 3 = 0 f_y=0=>2y-6x+3=0 f y = 0 => 2 y − 6 x + 3 = 0
x = − 111 16 , y = − 21 16 x=-{111 \over 16}, y=-{21 \over 16} x = − 16 111 , y = − 16 21
We find the critical point ( − 111 16 , − 21 16 ) (-{111 \over 16},-{21 \over 16}) ( − 16 111 , − 16 21 )
To find the nature of the critical point, we apply the second derivative test. We have
A = f x x = 2 , B = f x y = − 6 , C = f y y = 2 A=f_{xx}=2, B=f_{xy}=-6, C=f_{yy}=2 A = f xx = 2 , B = f x y = − 6 , C = f yy = 2
At the point ( − 111 16 , − 21 16 ) , (-{111 \over 16},-{21 \over 16}), ( − 16 111 , − 16 21 ) , we have
D = A C − B 2 = 2 ( 2 ) − ( − 6 ) 2 = − 32 < 0 D=AC-B^2 =2(2)-(-6)^2=-32<0 D = A C − B 2 = 2 ( 2 ) − ( − 6 ) 2 = − 32 < 0 Then the point ( − 111 16 , − 21 16 ) (-{111 \over 16},-{21 \over 16}) ( − 16 111 , − 16 21 ) is a saddle point.
(iv)Check whether the following functions are homogeneous or not.
(a)
f ( x , y ) = x y + 3 y 2 x + sin x y f(x, y)={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}} f ( x , y ) = y x + 2 x 3 y + sin y x f ( t x , t y ) = t x t y + 3 t y 2 t x + sin t x t y = f(tx, ty)={tx \over ty}+{3ty \over 2tx}+\sin{\sqrt{{tx \over ty}}}= f ( t x , t y ) = t y t x + 2 t x 3 t y + sin t y t x = = x y + 3 y 2 x + sin x y = t 0 f ( x , y ) ={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}}=t^0 f(x, y) = y x + 2 x 3 y + sin y x = t 0 f ( x , y ) A function f ( x , y ) f(x,y) f ( x , y ) is homogeneous.
(b)
f ( x , y ) = x 4 + 4 x 2 + y 2 x 2 f(x, y)=x^4+4x^2+{y^2 \over x^2} f ( x , y ) = x 4 + 4 x 2 + x 2 y 2 f ( t x , t y ) = ( t x ) 4 + 4 ( t x ) 2 + ( t y ) 2 ( t x ) 2 = t 4 x 4 + 4 t 2 x 2 + y 2 x 2 f(tx, ty)=(tx)^4+4(tx)^2+{(ty)^2 \over (tx)^2}=t^4x^4+4t^2x^2+{y^2 \over x^2} f ( t x , t y ) = ( t x ) 4 + 4 ( t x ) 2 + ( t x ) 2 ( t y ) 2 = t 4 x 4 + 4 t 2 x 2 + x 2 y 2 A function f ( x , y ) f(x,y) f ( x , y ) is not homogeneous.
(v)
f ( x , y ) = x 5 + 10 x 3 y 3 + 8 y 4 f(x,y)=x^5+10x^3y^3+8y^4 f ( x , y ) = x 5 + 10 x 3 y 3 + 8 y 4 Evaluate f x y f_{xy} f x y at a point ( x , y ) (x,y) ( x , y )
f x ( x , y ) = 5 x 4 + 30 x 2 y 3 f_x(x,y)=5x^4+30x^2y^3 f x ( x , y ) = 5 x 4 + 30 x 2 y 3 f x y ( x , y ) = 90 x 2 y 2 f_{xy}(x,y)=90x^2y^2 f x y ( x , y ) = 90 x 2 y 2 Verify that the function f f f satisfies the requirements of Schwarz’s theorem and hence evaluate f y x f_{yx} f y x
Since 90 x 2 y 2 90x^2y^2 90 x 2 y 2 is a polynomial, f x y ( x , y ) f_{xy}(x,y) f x y ( x , y ) is a continuous function.
Further, f y = 30 x 3 y 2 + 32 y 3 f_y=30x^3y^2+32y^3 f y = 30 x 3 y 2 + 32 y 3 exists. Hence f f f satisfies the conditions of Schwarz's theorem and so
f y x ( x , y ) = f x y ( x , y ) = 90 x 2 y 2 f_{yx}(x,y)=f_{xy}(x,y)=90x^2y^2 f y x ( x , y ) = f x y ( x , y ) = 90 x 2 y 2
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