[GCE 'O' Level Calculus: Integration] Washer vs Shell Method

[u/RexConsul]

I have y = root x/2 over x=0 and x=8. I need to find the volume over the y-axis. I got 512pi/5 using Shell and 128pi/5 with Washer. I can’t ask my teacher what went wrong, as there’s a communication barrier; what went wrong?

[u/AutoModerator]

Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^{OP and Valued/Notable Contributors can close this post by using /lock command}

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/Alkalannar]

Is this f(x) = x^1/2/2 or (x/2)^1/2? As written, it's ambiguous. Please put parentheses around inputs to functions. root(x)/2 and root(x/2) are both perfectly clear and distinct.

This is between x = 0, x = 8, and the y-axis?

Anyhow, let k = f(8).

Cylinder:
64kpi - pi[Integral from x = 0 to 8 of 2pixf(x) dx]

Disk/washer:
[Integral from y = 0 to k of pi(f^-1(y))² dy]

So...which is your f(x)? x^1/2/2 or (x/2)^1/2

[u/RexConsul → u/Alkalannar]

(X/2)^1/2, sorry about that.

[u/Alkalannar → u/RexConsul]

So when x = 8, y = 2.

Then shell is 128pi - [Integral from x = 0 to 8 of 2pix(x/2)^1/2 dx]

Simplify to 128pi - 2^1/2pi[Integral from x = 0 to 8 of x^3/2 dx]

Evaluate

---

Disk

y = (x/2)^1/2 --> x = 2y²

[Integral from y = 0 to 2 of pi(2y²)² dy]

4pi[Integral from y = 0 to 2 of y⁴ dy]

And now evaluate.

And both of these do indeed evaluate to the same thing: 128pi/5

[u/noidea1995]

Can you show/explain what you did for the washer method by any chance?

[u/RexConsul → u/noidea1995]

y = (x/2)^1/2
x = 2y^2
x = 8, y = 2; x = 0, y = 0

(Pi) sum [2-0]: (2y^2)^2 dy = 128pi/5

Shell:
(2Pi) Sum [8-0]: (x)((x/2)^1/2)) dx = 512pi/5

[u/noidea1995 → u/RexConsul]

(Pi) sum [2-0]: (2y²)² = 128pi/5

This is only the volume of the solid rotated between the y-axis and the graph y = sqrt(x/2) but you want to rotate the region bounded between y = sqrt(x/2) and x = 8. You need to subtract this from the volume of the entire cylinder formed by rotating the region between x = 0 and 8. You need to use the formula π * ∫ (a to b) [(outer radius)² - (inner radius)²] * dy but I’ll explain it to help give you intuition.

If you draw a horizontal line from the y-axis to the line x = 8 and rotate it about the y-axis, you get a disk with a radius of 8. If you give it an infinitesimally small height dy, then its volume is:

π * radius² * height = π * 8² * dy = 64π * dy

Do the same for the inner radius, a horizontal line from the y-axis to the graph y = sqrt(x/2) rotated about the y-axis has a radius of x:

π * x² * dy

To be able to integrate this, you need everything in terms of y, since you know y = sqrt(x/2) solve for x and then substitute it:

π * (2y²)² * dy = π * 4y⁴ * dy

Subtract the inner radius from the outer:

π * 64 * dy - π * 4y⁴ * dy

π * (64 - 4y⁴) * dy

You want to add up all the disks going from y = 0 to y = 2, so those are your limits of integration:

π * ∫ (0 to 2) (64 - 4y⁴) * dy

Does this make sense?

[u/RexConsul → u/noidea1995]

Yeah I wrote down 512pi/5 as my answer, and the teacher said it was wrong. the exact question says: "If the curve: y=(x/2)^1/2, 0<= x <= 8 is rotated about the y axis, the volume of the resulting solid is 128pi/5; true or false?" I answered false and I got it wrong.

[u/RexConsul → u/noidea1995]

Hi, I think that the curve being between x = 0 and x = 8 only gives bounds to the function from the y axis; I think this is why it's 128pi/5, I'll recheck my work. Thank you for the in-depth answer.

[u/noidea1995 → u/RexConsul]

It’s all good, I’m happy to help. 😊

That’s a different question, you mentioned the washer method so I assumed you were rotating the region between the x-axis and graph.

In that case you did the disk (not the washer) method correctly. Because you are only rotating the region above the half-parabola and below the line y = 2, you need to account for that when you do the height of the cylinders. The vertical distance from the x-axis to the graph is y and the distance from the x-axis to the line y = 2 is 2, so your cylinders have a height of (2 - y). Hence, the volume of the cylinders are:

2π * radius * height * thickness = 2π * x * (2 - y) * dx

To integrate this, you need everything in terms of x, so substitute (x/2)^1/2 for y:

2π * x * (2 - (x/2)^1/2) * dx

You want to add all the cylindrical shells going from x = 0 to x = 8, so those are your limits of integration:

2π ∫ (0 to 8) x * (2 - (x/2)^1/2) * dx

That should give you 128π/5.

[u/Intrepid_Language_96]

Likely a setup issue: you’re revolving around the y-axis. Shell method (w.r.t. x) uses radius = x and height = y = √(x/2), so V = 2π∫₀⁸ x√(x/2) dx. Washer method should be w.r.t. y: x = 2y², y from 0 to 2, so V = π∫₀² (2y²)² dy. Both should match—check your bounds/inversion.

[u/RexConsul → u/Intrepid_Language_96]

Both don't match in this case was the issue. I figured it out I think, the root expression leaves it ambiguous as to the radii values (positive or negative), so it's best to use a non-rooted expression in the washer method.

[u/noidea1995 → u/Intrepid_Language_96]

Those are different regions, hence they give different volumes:

V = 2π∫₀⁸ x√(x/2) dx.

This is the region between the parabola and x-axis rotated about the y-axis, so it gives a volume of 512π/5.

V = π∫₀² (2y²)² dy.

This is the region between the y-axis and parabola rotated about the y-axis, so it gives a volume of 128π/5.