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.
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!
3 answers
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.
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.