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

#1**+10 **

Best Answer

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

The formula for the surface area (including the base) is pi*r^{2} + pi*r*s where s is the slant height (= √(h^{2} + r^{2}).

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

I'll let you do the numerics.

.

Alan
Mar 8, 2015

#2**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**+5 **

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

Note that A is written as 2*pi*sqrt(1+(h/r)^{2})*r^{2}/2 = pi*sqrt(r^{2} + h^{2})*r = pi*s*r

You must use ds, not just dx or dy.

.

Alan
Mar 10, 2015

#5**+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