Question

How do you find the surface area of the part of the circular paraboloid $ z = {x^2} + {y^2} $ that lies inside the cylinder $ {x^2} + {y^2} = 1 $ ?

Answer 1

Hint: $ z = f(x,y) $ defines a three-dimensional surface in XYZ-plane, to find the area of a shape in two dimensions, we use simple integration but to find the area of a three-dimensional surface, we use the concept of double integrals. In the given question, we have to find the area bounded by a paraboloid and cylinder, both the shapes are three dimensional so we use double integral in this question.

Complete step-by-step answer:
The area of a surface $ f(x,y) $ above a region R of the XY-plane is given by $ \iint\limits_R {\sqrt {{{(f'x)}^2} + {{(f'y)}^2} + 1} }dxdy $
 $ f'x $ and $ f'y $ are the partial derivatives of $ f(x,y) $ with respect to x and y respectively.
We are given that $ z = {x^2} + {y^2} \Rightarrow f(x,y) = {x^2} + {y^2} $
So, $ f'x = 2x $ and $ f'y = 2y $
For the region defined by $ {x^2} + {y^2} = 1 $ , the surface is given as –
 $
  S = \iint\limits_R {\sqrt {{{(2x)}^2} + {{(2y)}^2}} }dxdy \\
   \Rightarrow S = \iint\limits_R {\sqrt {4{x^2} + 4{y^2}} dxdy} \;
  $
But in the given double integral we have two different variables, so to make the integration easier; we convert rectangular coordinates into a function of polar coordinates: $ dxdy \to (r)drd\theta $
Using this information in the double integral, we get –
 $
  S = \int\limits_{\theta = 0}^{2\pi } {\int\limits_{r = 0}^1 {{{(4{r^2} + 1)}^{\dfrac{1}{2}}}} (r)drd\theta } \\
   \Rightarrow S = \int\limits_{\theta = 0}^{2\pi } {[\dfrac{{{{(4{r^2} + 1)}^{\dfrac{3}{2}}}}}{{12}}]_{r = 0}^1d\theta } \\
   \Rightarrow S = \int\limits_{\theta = 0}^{2\pi } {\dfrac{{\left( {5\sqrt 5 - 1} \right)}}{{12}}d\theta } \\
   \Rightarrow S = \dfrac{{5\sqrt 5 - 1}}{{12}}[\theta ]_{\theta = 0}^{2\pi } \\
   \Rightarrow S = \dfrac{{5\sqrt 5 - 1}}{{12}} \times 2\pi \\
   \Rightarrow S = \dfrac{{5\sqrt 5 - 1}}{6}\pi \;
  $
Hence, the surface area of the part of the circular paraboloid $ z = {x^2} + {y^2} $ that lies inside the cylinder $ {x^2} + {y^2} = 1 $ is equal to $ \dfrac{{5\sqrt 5 - 1}}{6}\pi $ square units.
So, the correct answer is “$ \dfrac{{5\sqrt 5 - 1}}{6}\pi $ square units.”.

Note: A parabola is an open curve; all the points lying on a parabola are equidistant from a fixed point (called the focus) and a fixed-line (called the directrix). A similar curve is followed by a projectile under the influence of gravity. There is a line that goes from the middle of a parabola about which the parabola is symmetrical, that is the line divides the parabola into two parts of similar shapes, this line is called the axis of the parabola. When a parabola is rotated about its axis, the three-dimensional figure obtained is called a paraboloid.