Volume: Revolving Region Bounded by Equations - Homework Solution

[clairez93]

Homework Statement

1. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

$$y = \sqrt{x}, y = 0, x = 4$$

the line x = 62. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=4.

$$y = \frac{1}{x}, y = 0, x = 1, x = 4$$

Homework Equations

The Attempt at a Solution

1.

$$V = \pi \int^{\sqrt{6}}_{0} ((6-y^{2})^{2} - (6-4)^{2}) dy$$
$$= \pi \int^{\sqrt{6}}_{0} (36 - 12y^{2} + y^{4} -4) dy$$
$$= \pi \int^{\sqrt{6}}_{0} (32 - 12y^{2} + y^{4}) dy$$
$$= \pi [32y - 4y^{3} + \frac{y^{5}}{5}]^{\sqrt{6}}_{0}$$
$$= \frac{76\sqrt{6}\pi}{5}$$

I am relatively sure that I integrated the expression I set up correctly, because I checked it with my ti-89. That said, that means the original integral I set up is incorrect. I'm not sure why.

2.

$$V = \pi \int^{4}_{1} (4^{2} - \frac{1}{x^{2}}) dx$$
$$= \pi \int^{4}_{1} (16 - \frac{1}{x^{2}}) dx$$
$$= \pi [16x + \frac{1}{x}]^{4}_{1}$$
$$= \pi [(64 + \frac{1}{4)}) - (16+1)]$$
$$= \pi (\frac{257}{4} - 17)$$
$$= \frac{189\pi}{4}$$

Again, checked with my calculator and I evaluated integral correctly. Set up went wrong.

 

[foxjwill]

For the first one, check the limits of integration. What values of y does the region cover?

EDIT: I can't see anything wrong at first glance with the second one. What is the answer you were given?

 

[clairez93]

Answers for problems:

1. $$\frac{192 \pi}{5}$$
2. $$\pi (8 ln 4 - \frac{3}{4}) = 32.49$$

Oh, should the upper limit be 2?

 

[foxjwill]

Oh, should the upper limit be 2?

Yep!

And as for the other one, it should be (4-1/x)^2, not 1/x^2.

EDIT: As a quick LaTeX hint, type 'ln' as '\ln' as in $$\ln x$$.