another problem :/

Using the CYLINDRICAL SHELL METHOD find the volume generated by rotating the region bounded by $x = 1 + (y - 2)^2$ and x = 2 about the x-axis.

I got $\frac{32\pi}{3}$ but the book says it's $\frac{16\pi}{3}$

Question

another problem :/

Using the CYLINDRICAL SHELL METHOD find the volume generated by rotating the region bounded by $x = 1 + (y - 2)^2$ and x = 2 about the x-axis.

I got $\frac{32\pi}{3}$ but the book says it's $\frac{16\pi}{3}$

lgbasallote · Answer

$$2\pi \int_1^3 y(5 + y^2 - 4y) dy$$ this is the equation i used

lgbasallote · Answer

@apoorvk ? can you help mate?

lgbasallote · Answer

can you help me @SmoothMath this has been hanging out here long :/

marco26 · Answer

$$2pi \int\limits_{1}^{3}y(-y^2+4y-3)dy$$

lgbasallote · Answer

why -3? @marco26 \(-2)^2 = 4

lgbasallote · Answer

$(-2)^2 = 4$ *

lgbasallote · Answer

unless you took out a negative...

lgbasallote · Answer

which seems not to be the case

marco26 · Answer

Here's my equation:$$2pi \int\limits_{1}^{3}y[2-(y^2-4y+5)]$$

lgbasallote · Answer

where did you get 2 -(y^2 - 4y +5)?

marco26 · Answer

the formula for horizontal strip for shell method is$$V=2pi \int\limits_{y1}^{y2}y_{c}(x_{R}-x_{L})dy$$
where yc is the distance of the element from the axis of revolution.

In your problem, Xr=2, and Xl=y^2+4y-5, and yc=y,, just plug them in

marco26 · Answer

oops i mean Xl= y^2-4y+5

lgbasallote · Answer

what does r and l mean?

lgbasallote · Answer

ohhh this is the area formula right??

marco26 · Answer

no, this is the formula for volume using shell method, given you have horizontal strip

marco26 · Answer

X=Xr-Xl

lgbasallote · Answer

no..i meant xr - xl that's area bounded between two curves or something

marco26 · Answer

here,XR-XL will be the height of the cylinder, yc is its radius, dy is the thickness

lgbasallote · Answer

uhmm you didnt really answer what i asked o.O

anonymous · Answer

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