The number of elements in the relation $$R= \left\{(x,y): 4x^{2}+y^{2}<52,x,y\in Z\right\}$$ is

We have the relation, $$R= \left\{(x,y): 4x^{2}+y^{2}<52,x,y\in Z\right\}$$

where, $$4x^2+y^2<52$$ gives

$$\Rightarrow y^2< 52-4x^2$$

$$\Rightarrow y<\sqrt{52-4x^2}$$  or  $$y>-\sqrt{52-4x^2}$$

Since $$\sqrt{k}$$ has to have $$k\geq 0$$, we find that $$4x^2$$ has to be less than or equal to $$52$$, i.e. $$x^2<13$$

Thus, for integral $$x$$, the values of $$x$$ can be $$-3, -2, -1, 0, 1, 2, 3$$

For $$x=-3$$ and $$x=3$$, we get $$y<\sqrt{52-36}$$  or  $$y>-\sqrt{52-36}$$ which gives $$y<4$$ or $$y>-4$$, and hence $$7$$ values that satisfy. Giving a total of $$7\times 2= 14$$ cases.

For $$x=-2$$ and $$x=2$$, we get $$y<\sqrt{52-16}$$ or $$y>-\sqrt{52-16}$$ which gives $$y<6$$ or $$y>-6$$, and hence $$11$$ values that satisfy. Giving a total of $$11\times 2= 22$$ cases.

For $$x=-1$$ and $$x=1$$, we get $$y<\sqrt{52-4}$$ or $$y>-\sqrt{52-4}$$ which gives $$y<4\sqrt{3}$$ or $$y>-4\sqrt{3}$$, and hence $$13$$ values that satisfy. Giving a total of $$13\times 2= 26$$ cases.

For $$x=0$$, we get $$y<\sqrt{52-0}$$ or $$y>-\sqrt{52-0}$$ which gives $$y<7.21$$ or $$y>-7.21$$, and hence $$15$$ values that satisfy. Giving a total of $$15$$ cases.

Therefore, we have $$14+22+26+15=77$$ cases that satisfy.