# Thread: Need help with Volume of 2 Curves about x = 4

1. ## Need help with Volume of 2 Curves about x = 4

My two equations are y=10ln(x) and y= -x^4 - x + 4. They are bounded below by the x axis. I need to find the volume of the shape obtained by revolving them around the line x=4. I'm not sure what method I should use to solve this, or how I would set it up. Could somebody please help me with this?

2. ## Re: Need help with Volume of 2 Curves about x = 4

Originally Posted by TheNerdyGinger
My two equations are y=10ln(x) and y= -x^4 - x + 4. They are bounded below by the x axis. I need to find the volume of the shape obtained by revolving them around the line x=4. I'm not sure what method I should use to solve this, or how I would set it up. Could somebody please help me with this?
Please check your check your class-notes and/or your textbook for corrections to this post.
If you look at this plot, surely you will agree that the question as stated is ridiculous.

3. ## Re: Need help with Volume of 2 Curves about x = 4

Have you graphed the region to be rotated? Using the disk method, each "disk", at a value of y, will have radius equal to 4 minus the x value corresponding to that y. For y from 0 up to the y-value where the two curves cross, the radius will be $4- (4- x- x^4)= x^4+ x$. From the y-value where the two curves cross, the radius will be 4- (10 ln(x)). The crucial question is "where do those two curves cross?". Can you solve $10ln(x)= 4- x- x^2$?

4. ## Re: Need help with Volume of 2 Curves about x = 4

Originally Posted by HallsofIvy
Have you graphed the region to be rotated? Using the disk method, each "disk", at a value of y, will have radius equal to 4 minus the x value corresponding to that y. For y from 0 up to the y-value where the two curves cross, the radius will be $4- (4- x- x^4)= x^4+ x$. From the y-value where the two curves cross, the radius will be 4- (10 ln(x)). The crucial question is "where do those two curves cross?". Can you solve $10ln(x)= 4- x- x^2$?
In the plot can the be bounded below by the x-axis?

5. ## Re: Need help with Volume of 2 Curves about x = 4

It's just one of those self imposed restrictions I guess.

6. ## Re: Need help with Volume of 2 Curves about x = 4

Are you sure it would be the disk method? Because there would be that space in between them since it's rotated around x = 4.

7. ## Re: Need help with Volume of 2 Curves about x = 4

I believe the region to be rotated is small triangular-shaped as shown in the attached graph. Use of cylindrical shells w/respect to x ... a calculator is necessary to determine the intersection of both curves and one x-intercept, not to mention calculating the volume.

$\displaystyle V=2\pi \int_1^{1.1311} (4-x) \cdot 10\ln{x} \, dx + 2\pi \int_{1.1311}^{1.2838} (4-x) \cdot (4-x-x^4) \, dx$

8. ## Re: Need help with Volume of 2 Curves about x = 4

That one makes more sense. Thank you!

9. ## Re: Need help with Volume of 2 Curves about x = 4

Another question: How would you set it up if I were to revolve those same equations around Y=4? Wouldn't I need to rewrite the equations in terms of y?

10. ## Re: Need help with Volume of 2 Curves about x = 4

Originally Posted by TheNerdyGinger
Another question: How would you set it up if I were to revolve those same equations around Y=4? Wouldn't I need to rewrite the equations in terms of y?
washers w/respect to x

two integrals again ...

$\displaystyle V = \pi \int_1^{1.1311} 4^2 - (4-10\ln{x})^2 \, dx + \pi \int_{1.1311}^{1.2838} 4^2 - (4-x-x^4)^2 \, dx$

11. ## Re: Need help with Volume of 2 Curves about x = 4

I guess I over thought that one... I'm starting to understand your tagline.

Thanks again guys!