1. ## work with integral

a.

If a force of 90 N stretches a spring 1 m beyond its natural length. how much work does it take to stretch the spring 5 m beyond its natural length?

b.

to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??

a.
1125 joules

i am at a utter lost for words.

3. Originally Posted by Legendsn3verdie

a.

If a force of 90 N stretches a spring 1 m beyond its natural length. how much work does it take to stretch the spring 5 m beyond its natural length?

This is an application of Hookes law that says $F=kx$.

We're initially told that it takes a force of 90 N to stretch it 1 m. This gives us enough info to find k, the spring constant:

$90=k(1)\implies k=90$

Thus, $F=90x$.

Therefore, the total amount of work done to stretch it to 5 m is $W=90\int_0^5 x\,dx$

Can you take it from here?

b.

to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??
I don't have much time to do this one. I will come back later, if no one else has answered this.

--Chris

4. hm still need help on b, really strugglin.

5. Demonstrate your efforts, please. What have you tried? What did the integral look like? Why do you think it is incorrect? That is an awesome problem! You can learn a ton of calculus by solving it. Go!!

6. Originally Posted by TKHunny
Demonstrate your efforts, please. What have you tried? What did the integral look like? Why do you think it is incorrect? That is an awesome problem! You can learn a ton of calculus by solving it. Go!!
here is what i did for b.. the right answer is 21 mil, i got 17mil

to design the interior surface of a huge staiinless-steel tank, you revolve the curve y = x^2, 0 <= x <= 4, about the y axis. the container with dimensions in meters, is to be filled with seawater, which weighs 10,000N/m^3. how much work will it take to empty the tank by pumping the water to the tanks top??

anyone have a clue??

7. Take a good hard look at "16-x" vs your limits "[0,4]". You've pumped from 12 to 16. What about the rest?

Personally, I would try this:

$\int_{0}^{16}\left[\pi x^{2}\right]\cdot (16 - y)\;dy\;=\;\int_{0}^{16}[\pi y] \cdot (16 - y)\;dy$