The region is just 1/2 of one arch of the curve
using discs of thickness dx,
v = ∫[0,π/14] π(R^2-r^2) dx
where R = 8+cos7x and r = 8
v = ∫[0,π/14] π((8+cos(7x))^2-8^2) dx = π/28 (π+64)
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y=0, y=cos(7x) , x=π/14, x=0 about the axis y=−8
5 answers
What about this one:
Find the volume formed by rotating the region enclosed by:
x=5.5y and y^3=x with y≥0 about the y-axis
I'm just confused on which formula I should be using to solve these problems
Find the volume formed by rotating the region enclosed by:
x=5.5y and y^3=x with y≥0 about the y-axis
I'm just confused on which formula I should be using to solve these problems
there are only two formulas.
discs: v = πr^2 dx (or dy)
shells: v = 2πrh dx (or dy)
Sometimes one is more convenient than the other, depending on the boundary of the region. Always sketch the region to get a feel for it.
For this one, either one works.
find where the curves intersect. The region is convex, so having to shift boundaries will not be a problem. There is a problem almost identical to this at
https://www.jiskha.com/questions/1837054/find-the-volume-formed-by-rotating-about-the-y-axis-the-region-enclosed-by-x-11y-and-y-3-x
the answer is kinda messy, though.
discs: v = πr^2 dx (or dy)
shells: v = 2πrh dx (or dy)
Sometimes one is more convenient than the other, depending on the boundary of the region. Always sketch the region to get a feel for it.
For this one, either one works.
find where the curves intersect. The region is convex, so having to shift boundaries will not be a problem. There is a problem almost identical to this at
https://www.jiskha.com/questions/1837054/find-the-volume-formed-by-rotating-about-the-y-axis-the-region-enclosed-by-x-11y-and-y-3-x
the answer is kinda messy, though.
could you use either equation for all problems ?
yes, but you may have to split it up into more than one part as you try to avoid duplicating bits. Consider the region bounded by
y = e^x, y = e^-x and the line x=1.
If you rotate that around the x-axis using washers (discs with holes)
v = ∫[0,1] π((e^x)^2 - (e^-x)^2) dx
Using shells, you have to change boundaries at y=1
v = ∫[1/e,1] 2πy(1-ln(1/y)) dy + ∫[1,e] 2πy(1-lny) dy
To rotate around the y-axis, things change. Using shells,
v = ∫[0,1] 2πx(e^x - e^-x) dx
Using washers,
v = ∫[1/e,1] π(1^2 - (ln(1/y))^2) dy + ∫[1,e] π(1^2 - (lny)^2) dy
y = e^x, y = e^-x and the line x=1.
If you rotate that around the x-axis using washers (discs with holes)
v = ∫[0,1] π((e^x)^2 - (e^-x)^2) dx
Using shells, you have to change boundaries at y=1
v = ∫[1/e,1] 2πy(1-ln(1/y)) dy + ∫[1,e] 2πy(1-lny) dy
To rotate around the y-axis, things change. Using shells,
v = ∫[0,1] 2πx(e^x - e^-x) dx
Using washers,
v = ∫[1/e,1] π(1^2 - (ln(1/y))^2) dy + ∫[1,e] π(1^2 - (lny)^2) dy
1 answer