Solveeit Logo
Login

Question

Question: How do you find the surface area of the part of the circular paraboloid \( z = {x^2} + {y^2} \) that...

How do you find the surface area of the part of the circular paraboloid z=x2+y2z = {x^2} + {y^2} that lies inside the cylinder x2+y2=1{x^2} + {y^2} = 1 ?

Explanation

Solution

Hint : z=f(x,y)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)f(x,y) above a region R of the XY-plane is given by ∬R(f′x)2+(f′y)2+1dxdy\iint\limits_R {\sqrt {{{(f'x)}^2} + {{(f'y)}^2} + 1} }dxdy
f′xf'x and f′yf'y are the partial derivatives of f(x,y)f(x,y) with respect to x and y respectively.
We are given that z=x2+y2⇒f(x,y)=x2+y2z = {x^2} + {y^2} \Rightarrow f(x,y) = {x^2} + {y^2}
So, f′x=2xf'x = 2x and f′y=2yf'y = 2y
For the region defined by x2+y2=1{x^2} + {y^2} = 1 , the surface is given as –
S=∬R(2x)2+(2y)2dxdy ⇒S=∬R4x2+4y2dxdy   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→(r)drdθdxdy \to (r)drd\theta
Using this information in the double integral, we get –
S=∫θ=02π∫r=01(4r2+1)12(r)drdθ ⇒S=∫θ=02π[(4r2+1)3212]r=01dθ ⇒S=∫θ=02π(55−1)12dθ ⇒S=55−112[θ]θ=02π ⇒S=55−112×2π ⇒S=55−16π   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=x2+y2z = {x^2} + {y^2} that lies inside the cylinder x2+y2=1{x^2} + {y^2} = 1 is equal to 55−16π\dfrac{{5\sqrt 5 - 1}}{6}\pi square units.
So, the correct answer is “55−16π\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.