Finding the equations of surfaces of revolution

Question

I have the following question:

$$\text{Sketch and find the equations of the surfaces formed by}$$ $$\text{i) }x^2 - y^2 + 1 = 0 \text{ about the y-axis}$$ $$\text{ii) }x^2 - 2y^2 + 2a^2 = 0 \text{ about the x-axis}$$

What's a general method to answer these kinds of questions?

Thanks

Answer 1

For revolution of the curve $$\text{i) }x^2 - y^2 + 1 = 0 \text{ about the y-axis}$$

You may express $$x(y)=\sqrt{1-y^2}$$ then $$X(y,\phi)=\sqrt{1-y^2}\cos\phi$$ $$Y(y,\phi)=\sqrt{1-y^2}\sin\phi$$

Will be the equation for the surface formed by revolution. Notice that $X(y,0)=x(y)$.

For revolution of the curve $$\text{ii) }x^2 - 2y^2 + 2a^2 = 0 \text{ about the x-axis}$$ You may express $$y(x)=\sqrt{a^2+(x^2)/2}$$ then $$X(x,\phi)=\sqrt{a^2+(x^2)/2}\sin\phi$$ $$Y(x,\phi)=\sqrt{a^2+(x^2)/2}\cos\phi$$

Will be the equation for the surface formed by revolution. Notice that $Y(x,0)=y(x)$.

Answer 2

In the equation $x^2-y^2+1=0$, you can simply put $x^2+z^2$ instead of $x^2$ to get $$x^2+z^2-y^2+1=0.$$ The reason is that $r^2=x^2+z^2$ expresses the distance $r$ between a point and the $y$-axis. When a point is rotated around $y$-axis, the distance of the point from the axis remains unchanged.