Volumes of revolution

[Vader07]

Hello everyone,


Here is the problem

Find the volume of the solid of revolution formed by rotating the region bounded by the x-axis and the graph of

y = x^3 square root (5 + sin (pi)(x^7)

from x =0 to x = 1 about the x-axis


This is what I did so far:

3x + (5 + cos 7x)
3 + (5 + cos 7x)
I am not sure if this is the right way to do, so any help would be greatly appreciated

 

[galactus]

Please brush up on your notation. What is it you mean.

\(\displaystyle f(x)=x^{3}\sqrt{5+x^{7}sin({\pi})}\)?.

Whatever it is, try using washers.

\(\displaystyle {\pi}\int_{0}^{1}[f(x)]^{2}dx\)

 

[Vader07]

Dear galactus,



The equation you showed me is correct, sorry for the confusion.
Can you please show me the first three steps of how to do this problem
I am completely lost in this problem

Thanks in advance.

 

[galactus]

Completely lost?. Don't be. Just use a little algebra to break it down to something easier to deal with.

Washers:

\(\displaystyle {\pi}\int_{0}^{1}\left(x^{3}\sqrt{5+x^{7}sin({\pi})}\right)^{2}dx={\pi}\int_{0}^{1}x^{6}(5+x^{7}sin(\pi))dx\)

\(\displaystyle {\pi}\left[5\int_{0}^{1}x^{6}dx+sin(\pi)\int_{0}^{1}x^{13}\right]dx\)

Now, integrate each and it should be rather easy, by comparison.

Remember, \(\displaystyle sin(\pi)=0\), so that makes it even easier. This problem looked much worse than it was.

If you feel brave, try using shells and see if you get the same result.