how to graph x^(2/3)+y^(2/3)=4^(2/3)

Question

I am trying to graph this before rotating around axes to find the volumes of solids

Answer 1

This is a pincushion. Find the intercepts first.

When x=0: y^⅔=4^⅔. Cube each side: y²=4², so y=±4. Intercepts at (0,-4) and (0,4).

When y=0, similarly, x=±4, intercepts at (-4,0) and (4,0). These 4 points are vertices. The figure is also symmetrical.

[x^⅔=4^⅔-y^⅔=(4^⅓+y^⅓)(4^⅓-y^⅓)=(∛4+∛y)(∛4-∛y).]

When y=1, we get x^⅔=4^⅔-1=1.52 approx. x²=1.52³=3.51, x=±√3.51=±1.87 approx.

Also, when x=1, y=±1.87. This gives you 4 more points to plot: (1,1.87), (1,-1.87), (1.87,1), (-1.87,1).

And, because of symmetry, you can also plot (-1.1.87), (-1,-1.87), (1.87,-1), (-1.87,-1).

From the 12 points you can sketch the graph.

Now you should be able to visualise the results of rotation to generate a solid.