Question

Revolve the figure shown about the y-axis completely to sweep out a solid of revolution.

Answer 1

Answer:

46.4 units^2

Step-by-step explanation:

Solution:-

- We will determine the surface area of revolution of figure given below about y-axis.

- We will compute using the calculus methods and verify by using basic area formula.

- The region of solid is bound by function f ( x ) and coordinate axes.

- The function can be expressed as a linear equation of line:

                              y = m*x + c

Where,

           m: Slope

           c : y-intercept

- The slope m is evaluated by using two pair of points. We will use the end points ( 2 , 0 )  and  ( 0 , 5 ) :                          

                            m = [ y2 - y1 ] / [ x2 - x1 ]

                            m = [ 5 - 0 ] / [ 0 - 2 ]

                            m = - 5 / 2 = -2.5

- The intercept is defined by coordinate ( 0 , 5 ). Where y = c = 5

- The linear function is:

                           y = -2.5x + 5

- From calculus the surface area of revolution of " S " about y-axis is given of the form:

                              $S = 2\pi \int\limits_S {x} \, ds$

Where, ds is the differential element of solid boundary defined by the bounding function f ( x ). The formulation of integral in x-coordinate system of ds is given as:

                            $ds = \sqrt{1 + (\frac{dy}{dx} )^2}.dx$

- Taking the first derivative of our bounding function y = f ( x ):

                            dy / dx = -2.5

- Substitute the "ds" length into the surface integral "S":

                             $ds = \sqrt{1 + (-2.5 )^2}.dx\\\\ds = \sqrt{1 + 6.25}.dx\\\\ds = \sqrt{7.25}.dx\\\\S = 2*\sqrt{7.25}*\pi\int\limits^2_0 {x} \, dx \\\\S = 2*\sqrt{7.25}*\pi*\frac{x^2}{2} \\\\S = 2*\sqrt{7.25}*\pi*\frac{2^2}{2} = 4*\pi*\sqrt{7.25}$

- The base area of the solid would be a circle of radius r = 2 units. The area of the circle Ac would be:

                            Ac = πr^2

                            Ac = π2^2 = 4π

-Hence,

The total surface area of the revolved region about y-axis would be:

                           Total Area = S + Ac

                           Total Area = 4π + 4π*√7.25

                          Total Area = 46.40236 units^2

Answer: The total area to the nearest tenth would be 46.4 unit^2.

                   

Answer 2

Answer:

46.37 square units

Step-by-step explanation:

If the right triangle is revolved the figure shown about the y-axis completely, the figure will be formed as a cone. As you can see in my attached photo.

So, the radius of the cone is : 2 units

the slant height is: $\sqrt{5^{2}+2^{2} }$ = $\sqrt{29}$ (use pytagon theory)

Hence,  the total surface area of the resulting figure

SA = π * the radius * the slant height + π*$radius^{2}$

= 3.14*2*$\sqrt{29}$ + 3.14*2*2

= 33.81 + 12.56

=46.37 square units