1. ## Curved surface area

I have the answers but i m unable to work it out

2. ## Re: Curved surface area

Originally Posted by haftakhan

I have the answers but i m unable to work it out
Hello,

the area in question is a rectangle with the length of 9.5 m and the width of a quarter perimeter of a circle with radius r = 0.8 m.

Therefore:

$\displaystyle A = 9.5\ m \cdot \frac14 \cdot 2 \cdot \pi \cdot 0.8 \ m \approx 11.94\ m^2$

3. ## Re: Curved surface area

What earboth calculated was the curved surface area of the cylinder, the area of contact with the ends of the cylinder must be added to this.
To find that area split the circle cross section up into three parts. 1. The area of contact, 2. The right angled triangle shown in the diagram, 3. The remainder of the circle which is 270 degrees around the centre.

4. ## Re: Curved surface area

Originally Posted by Shakarri
What earboth calculated was the curved surface area of the cylinder, the area of contact with the ends of the cylinder must be added to this.
To find that area split the circle cross section up into three parts. 1. The area of contact, 2. The right angled triangle shown in the diagram, 3. The remainder of the circle which is 270 degrees around the centre.
Hello,

thanks for pointing out that I forgot to calculate the front and rear part of the cross-sections. But: In this case the complete area of the cross-sections, which have contact to the oil, can be calculated as a half circle minus a half square:

$\displaystyle A_{complete}= \underbrace{9.5\ m \cdot \frac14 \cdot 2 \cdot \pi \cdot 0.8 \ m }_{\text{curved surface area}}+ \underbrace{\frac12 \cdot \pi \cdot 0.8^2}_{\text{half circle}} - \overbrace{\frac12 \cdot \frac12 \cdot 1.6^2}^{\text{half square, diameter as diagonal}} \approx 12.3\ m^2$