W05 - HW Solutions

01

(1) Integral formula:

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(2) Integrand components:

Width function:

Depth function:

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(3) Integrate:

02

(a) Left:

Set at the water line, increasing downwards.

Alternative: set at the top of the trapezoid. Obtain:

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(b) Center:

Set at the water line, increasing downwards.

Alternative: set at the top of the trapezoid. Obtain:

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(c) Right:

Set at the water line, increasing downwards.

03

(a) (1) Integral formula:

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(2) Setup:

Coordinate system: set at the top of the tank, increasing downwards.

Horizontal slice of the tank: disk of radius at depth , satisfies:

Distance pumped up (add for the spigot):

Thus:

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(b) (1) Change upper bound, top of water at :

04

(1) Integral formula:

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Option 1: (2) Setup:

Set at the bottom, increasing upwards.

Radius of the cone with a QLIF:

Horizontal slice of the cone tower: disk of radius at height , satisfies:

The slice at is raised a distance of .

Thus:

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Option 2: (2) Setup:

Set at the top of the cone, increasing downwards.

Now is the distance from the ground up to the height of a slice indexed by .

Radius function:

Thus:

05

(1) Integral formula:

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(2) Integrand components:

So we have:

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(3) Integrate:

(Assuming and .)

06

(1) Integral formula:

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Option 1:

(2) Using at water line, increasing downwards: (a) Left:

(b) Center:

(c) Right:

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Option 2:

(2) Using at top of shape, increasing downwards: (a) Left:

(b) Center:

(c) Right:

07

(1) Integral formula:

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Option 1:

(2) Using at water line, increasing downwards: (a) Left:

(b) Center:

(c) Right:

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Option 2:

(2) Using at center of shape, increasing downwards: (a) Left:

(b) Center:

(c) Right:

08

(1) Integral formula:

Let at the ground and increase going up.

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(2) Compute force:

The force on the rope (at the top) when the bucket is at height is:

We know .

Water is leaking at . Therefore:

The weight of chain remaining is:

Put together:

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(3) Integrate:

09

(1) Integral formula:

Set at the center of the tank.

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(a) (2) Geometry:

So:

Tank length is so a horizontal slice is a rectangle with area .

Depth is:

Therefore:

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(b) (2) Change bounds and height:

10

(1) Integral formula:

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(2) Integrand components:

Option 1: Set at the base, going up.

Take a cross-sectional slice with a vertical plane. This intersects the surface of the pyramid in a triangle whose width is the side length of the square (the horizontal cross section) at height .

Note that . So we have:

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Option 2:

Set at the vertex, going down.

And: