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Answer:
Step-by-step explanation:
Given that an engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min.
We convert everything in mm for uniformity
We have formula for volume as
Given that
h is constant
= 6954.67 mm/sec^3
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50