Question

(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)

Answer 1

(i)

$$\displaystyle\int_{0}^2\displaystyle\int_{x}^{2x}(x^4+y^2)dydx=\displaystyle\int_{0}^2\bigg[x^4y+{y^3 \over 3}\bigg]\begin{matrix} 2x \\ x \end{matrix}dx=$$ $$=\displaystyle\int_{0}^2\bigg(x^4(2x)+{(2x)^3 \over 3}-x^4(x)-{x^3 \over 3}\bigg)dx=$$ $$=\displaystyle\int_{0}^2\big(x^5+{7x^3 \over 3}\big)dx=\big[{x^6 \over6}+{7x^4 \over 12}\big]\begin{matrix} 2 \\ 0 \end{matrix}=$$ $$={2^6 \over 6}+{7(2)^4 \over 12}-0=20$$

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

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

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

$f_x=-2x, f_y=-2y$

$\sqrt{1+(f_x)^2+(f_y)^2}=\sqrt{1+(-2x)^2+(-2y)^2}=\sqrt{1+4x^2+4y^2}$

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

$\sqrt{1+4x^2+4y^2}=\sqrt{1+4r^2}$

Hence, the surface area $A$ is given by

$$A=\int\int_{E}\sqrt{1+(f_x)^2+(f_y)^2}dxdy=$$ $$=\displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^{5}\sqrt{1+4r^2}rdrd\theta=$$ $$=\displaystyle\int_{0}^{2\pi}\bigg[{(1+4r^2)^{3/2} \over 12}\bigg]\begin{matrix} 5 \\ 0 \end{matrix}d\theta=$$ $$={2\pi \over 12}\big((1+4(5)^2)^{3/2}-1\big)={\pi \over 6}\big((101)^{3/2}-1\big)$$

(iii)

$f(x,y)=x^2+y^2-6xy+6x+3y-4$

Partial derivatives

$f_x=2x-6y+6$

$f_y=2y-6x+3$

To find the critical points, we solve

$f_x=0=>2x-6y+6=0$

$f_y=0=>2y-6x+3=0$

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

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

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

$A=f_{xx}=2, B=f_{xy}=-6, C=f_{yy}=2$

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

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

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

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

(a)

$$f(x, y)={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}}$$ $$f(tx, ty)={tx \over ty}+{3ty \over 2tx}+\sin{\sqrt{{tx \over ty}}}=$$ $$={x \over y}+{3y \over 2x}+\sin{\sqrt{{x \over y}}}=t^0 f(x, y)$$

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

(b)

$$f(x, y)=x^4+4x^2+{y^2 \over x^2}$$ $$f(tx, ty)=(tx)^4+4(tx)^2+{(ty)^2 \over (tx)^2}=t^4x^4+4t^2x^2+{y^2 \over x^2}$$

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

(v)

$$f(x,y)=x^5+10x^3y^3+8y^4$$

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

$$f_x(x,y)=5x^4+30x^2y^3$$ $$f_{xy}(x,y)=90x^2y^2$$

Verify that the function $f$ satisfies the requirements of Schwarz’s theorem and hence evaluate $f_{yx}$

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

Further, $f_y=30x^3y^2+32y^3$ exists. Hence $f$ satisfies the conditions of Schwarz's theorem and so 

$$f_{yx}(x,y)=f_{xy}(x,y)=90x^2y^2$$