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a right-angled triangle ABC with AB=8cm and BC=10cm .A solid is formed by rotating triangle ABC about AB

describle the solid formed and find the total surface area of the solid in pi

Guest Mar 8, 2015

Best Answer 

 #1
avatar+27228 
+10

I've assumed that BC is not the hypotenuse. A cone is formed:

 cone

The formula for the surface area (including the base) is pi*r2 + pi*r*s where s is the slant height (= √(h2 + r2).  

 

Here: r = BC = 10cm, h = AB = 8cm

I'll let you do the numerics.

.

Alan  Mar 8, 2015
 #1
avatar+27228 
+10
Best Answer

I've assumed that BC is not the hypotenuse. A cone is formed:

 cone

The formula for the surface area (including the base) is pi*r2 + pi*r*s where s is the slant height (= √(h2 + r2).  

 

Here: r = BC = 10cm, h = AB = 8cm

I'll let you do the numerics.

.

Alan  Mar 8, 2015
 #2
avatar+94105 
0

Thanks Alan,

I think I am suffering from an embarrasing mental block.

C(10,0)     A(0,8)

equation AC:    

$$y=-0.8x+8$$

or

$$x= \frac{y-8}{-0.8}$$

---------------------------------------

 

For the SA of the curved surface why can't I just say

 

$$\\SA=\int_0^8\;\;2\pi x\;dy\\\\
SA=2\pi\int_0^8\;\;\frac{y-8}{-0.8}\;dy\\\\
SA=\frac{2\pi}{-0.8}\int_0^8\;\;y-8\;dy\\\\
SA=\frac{2\pi}{-0.8}\left[\;\;\frac{y^2}{2}-8y\right]_0^8\\\\\
SA=\frac{2\pi}{-0.8}\left[\frac{8^2}{2}-64\right]\\\\
SA=\frac{2\pi}{-0.8}\left[\frac{-62}{2}\right]\\\\
SA=80\pi\\\\$$

 

I know that this is wrong but I don't know why.

Melody  Mar 9, 2015
 #3
avatar+27228 
+5

This is how the surface area of the slanted surface should be calculated:

 

 surface area of cone

Note that A is written as 2*pi*sqrt(1+(h/r)2)*r2/2  =  pi*sqrt(r2 + h2)*r = pi*s*r

 

You must use ds, not just dx or dy. 

.

Alan  Mar 10, 2015
 #4
avatar+94105 
0

Thanks Alan

Melody  Mar 10, 2015
 #5
avatar+92674 
+5

Melody, we can also do this in this manner .

 

Surface area = 2pi ∫f(x) √[1 + [ f ' (x)]^2 ] dx  from a to b

If we rotate the line   f(x) = (10/8)x  = 1.25x  around the x axis  from x = 0 to x = 8 we have the same cone laid on its side..... and we have....  {Remember that f 'x  = 1.25.....and we must square this under the root }

2 pi ∫ (1.25x ) √[1 + [1.25]^2] dx   from  0 to 8 =

2pi √[1 + [1.25]^2] [ 1.25/2] x^2    from 0  to 8  =

 1.25pi √[1 + [1.25]^2] * 64  =  402.32 sq units

And we can add the area of the base to this.

 

  

CPhill  Mar 10, 2015
 #6
avatar+27228 
+5

Chris, I don't think you mean √[1 + [ f ' (x)]^2 ]^.5, but just √[1 + [ f ' (x)]^2 ] or [1 + [ f ' (x)]^2 ]^.5,

and this, multiplied by dx, is just ds!

.

Alan  Mar 10, 2015
 #7
avatar+92674 
+5

Thanks, Alan.....I forgot I already had the root in there....

 

  

CPhill  Mar 10, 2015
 #8
avatar+94105 
0

Thanks Alan and Chris,  I have to take time out to think about all this.    (◐‿◑)

Melody  Mar 11, 2015

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