Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve
is
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:

Where
,
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that
and the fact that
when
, the spring constant (
), measured in newtons per meter, is:



Now, the work function is obtained:

![W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%28250%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%5Cright%29%5Ccdot%20%5B%280.05%5C%2Cm%29%5E%7B2%7D-%280.00%5C%2Cm%29%5E%7B2%7D%5D)

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be
. The area of the region enclosed by one loop of the curve is given by the following integral:
![A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta](https://tex.z-dn.net/?f=A%20%3D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_0%20%7B%5Br%28%5Ctheta%29%5D%5E%7B2%7D%7D%20%5C%2C%20d%5Ctheta)

By using trigonometrical identities, the integral is further simplified:





The area of the region enclosed by one loop of the curve
is
.