Question

6) (15 points) A spherical storage tank extends from y=-4 to y=4. It is filled with a fluid weighing w lb per cubic foot. The equation of the circle is x+y=16 All of the fluid will be pumped up to the y=6 feet level. Calculate the work done. (Your answer will be an expression containing w.)

Answer 1

Consider a small disk of thickness dy at a height y as following:

-y=6 6-y dey EX CS Scanned with CamScanner

The raidus of this disk is x

But $x^2+y^2=16$ .

So x can be written as: $x=\sqrt{16-y^2}$ .

So the volume of the disk is:

.

V = arah = 7.2 dy = 7T T (16 – y?) dy

Thus weight of the liquid disk is:

$weight=w\cdot g \cdot \pi \left ( 16-y^2 \right )dy$

But g=32.So;'

$weight=32\pi w\cdot \left ( 16-y^2 \right )dy$

So, the work done to lift this disk of liquid to y=6ft level is:

$dW=\left (6-y \right )32\pi w\cdot \left ( 16-y^2 \right )dy$

Since y extends from -4 to 4; the total work done is:

$W=\int_{-4}^{4}\left (6-y \right )32\pi w\cdot \left ( 16-y^2 \right )dy$

$W=32\pi w\int_{-4}^{4}\left (6-y \right ) \left ( 16-y^2 \right )dy$

$W=32\pi w\int_{-4}^{4}\left ( 96-6y^2-16y+y^3 \right )dy$

$W=32\pi w\left [ 96y-2y^3-8y^2+\frac{y^4}{4} \right ]^4_{-4}$

$W=32\pi w\left ( 768-256-0+0 \right )$

$W=32\pi w\times 512$

$W=16384\,\pi\, w$.

Therefore the work done to lift entire fluid to y=6 level is $W=16384\,\pi\, w=51471.9\,w$ .

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