Asked by laura
a pie dish has base diameter 8 inches and top diameter 10 inches and height 2 inches.
what is the volume of the disk?
do I do it so cross sections are squares?
help!!
Thanks!
what is the volume of the disk?
do I do it so cross sections are squares?
help!!
Thanks!
Answers
Answered by
Reiny
I would visualize the pie dish as a cut-off part of a cone, which has been sliced horizontal to the base, 2 inches up from the base
let the distance to the vertex of the imaginary cone be x inches.
then by ratios:
5/(x+2) = 4/x
x = 8
volume of cone with height of 10 and diameter of 10 = 1/3(pi)(5^2)(10) = (250/3)pi
volume of the cone which is cut off
= 1/3pi(4^2)(8) = (128/3)pi
so the volume of the pie dish is
(250/3)pi - (128/3)pi = 122/3pi
= appr. 127.76 cu. in.
let the distance to the vertex of the imaginary cone be x inches.
then by ratios:
5/(x+2) = 4/x
x = 8
volume of cone with height of 10 and diameter of 10 = 1/3(pi)(5^2)(10) = (250/3)pi
volume of the cone which is cut off
= 1/3pi(4^2)(8) = (128/3)pi
so the volume of the pie dish is
(250/3)pi - (128/3)pi = 122/3pi
= appr. 127.76 cu. in.
Answered by
bobpursley
Let AreaBase, AreaTop be the area of the bottome and top. Then the area of the center (1/2 way up) is AreaMiddle
A very old approximating formula for volume of anything, even irregular objects, is
Volume= 1/6(areabottom+ 4AreaMiddle+ AreaTop)* height.
For instance, the area of a sphere: Areabottom, areatop is zero, area middle is PI r^2
Then the approximating formula is
Volume= 1/6(4PIr^2)2r= 4/3 PI r^3
That ought to put a kink in the profs text.
Now the calculus way:
Do it in cross sections. Relate r to height (r=4+h/2) so dr= dh/2
Then check it with that old approximating formula schoolboys memorized several hundred years ago.
Volume= INT PI(4+h/2)^2 dh
=INT PI (16+4h+h^2/4) dh
= PI(16h + 2h^2 + h^3/12 ) eval at the limits, or
Volume= PI(32+8+8/12)= 40.75PI by calculus, check my thinking.
Now the old approximating formula:
Volume= 1/6(PI)(16 + 4*20.25+ 25)2
1/6 PI (61.25)2=40.67 PI
That old formula is handy.
A very old approximating formula for volume of anything, even irregular objects, is
Volume= 1/6(areabottom+ 4AreaMiddle+ AreaTop)* height.
For instance, the area of a sphere: Areabottom, areatop is zero, area middle is PI r^2
Then the approximating formula is
Volume= 1/6(4PIr^2)2r= 4/3 PI r^3
That ought to put a kink in the profs text.
Now the calculus way:
Do it in cross sections. Relate r to height (r=4+h/2) so dr= dh/2
Then check it with that old approximating formula schoolboys memorized several hundred years ago.
Volume= INT PI(4+h/2)^2 dh
=INT PI (16+4h+h^2/4) dh
= PI(16h + 2h^2 + h^3/12 ) eval at the limits, or
Volume= PI(32+8+8/12)= 40.75PI by calculus, check my thinking.
Now the old approximating formula:
Volume= 1/6(PI)(16 + 4*20.25+ 25)2
1/6 PI (61.25)2=40.67 PI
That old formula is handy.
Answered by
Anonymous
Did you mean using shell method? Using shells is quite complicated for this problem :/.
Vertical
Height = x
Length = 2yPi = 2(-2x+10)Pi
Width = dx
Take the integral of 2x(-2x+10)Pidx from 4 to 5 and you get 26/3Pi. Then take the integral of 2(2xPi)dx from 0 to 4 and you get 32Pi. Add them together and you get 122/3Pi. Which is the same as the other answers.
Horizontal(took me well over 30 minutes -.- Didn't know I was so weak at shells..)
Height = 2-x
Length = 2Pi(.5x+4)
Width = dx
Integrate from 0 to 2 and add the volume of the inside cube which is 32Pi and you'll get the same answer.
Vertical
Height = x
Length = 2yPi = 2(-2x+10)Pi
Width = dx
Take the integral of 2x(-2x+10)Pidx from 4 to 5 and you get 26/3Pi. Then take the integral of 2(2xPi)dx from 0 to 4 and you get 32Pi. Add them together and you get 122/3Pi. Which is the same as the other answers.
Horizontal(took me well over 30 minutes -.- Didn't know I was so weak at shells..)
Height = 2-x
Length = 2Pi(.5x+4)
Width = dx
Integrate from 0 to 2 and add the volume of the inside cube which is 32Pi and you'll get the same answer.
There are no AI answers yet. The ability to request AI answers is coming soon!