Area, volume, and square cross-sections bounded by y = x³/(1+x²) and y = 4−2x

[kreil]

I have been having trouble with this kind of problem lately and I need to know if what I have done here is right (calculator problem but I have not evaluated anything yet):

Let R be the region bounded by the y-axis and the graphs of

$$y=\frac{x^3}{1+x^2}$$ and

$$y=4-2x$$

a) Find the area of R
b) Find the volume of the solid generated when R is revolved about the x-axis
c)The region R is the base of a solid. For this solid, each cross section parallel to the x-axis is a square. Find the volume of this solid.


a)
a=point of intersection of the two graphs

$$\int_0^a(4-2x-\frac{x^3}{1+x^2})dx$$

b)
$${\pi}\int_0^a(4-2x)^2-(\frac{x^3}{1+x^2})^2dx$$

c)
$$\pi \int_0^a(4-2x-\frac{x^3}{1+x^2})^2dx$$

 

[UrbanXrisis]

there shouldn't be a $\pi$ in front of the integral on letter c.

everything else is fine

 

[thepatientmental]

First of all, it is great that you are seeking help and checking your work. It is important to make sure your solutions are correct when solving calculus problems.

For part a), finding the area of R, you correctly set up the integral using the bounds of 0 and a, and the two given functions. However, there is a small error in your integrand. Remember that when finding the area between two curves, the top function should be subtracted by the bottom function. So the correct integral should be:

\int_0^a(4-2x)-(\frac{x^3}{1+x^2})dx

For part b), finding the volume of the solid generated when R is revolved about the x-axis, you correctly set up the integral using the bounds of 0 and a, and the two given functions. However, the integrand should be the volume of a disc, which is {\pi}r^2, where r is the distance from the function to the axis of rotation. So the correct integral should be:

{\pi}\int_0^a[(4-2x)-(\frac{x^3}{1+x^2})]^2dx

For part c), finding the volume of the solid where each cross section parallel to the x-axis is a square, you correctly set up the integral using the bounds of 0 and a, and the two given functions. However, the integrand should be the area of a square, which is (side length)^2. In this case, the side length is the difference between the two functions. So the correct integral should be:

\int_0^a[(4-2x)-(\frac{x^3}{1+x^2})]^2dx

Overall, your setup and approach to solving these problems is correct. Just make sure to pay attention to the integrand and the difference between finding area and volume. Keep practicing and you will continue to improve. Good luck!