More volume-of-solid problems: x=y^2, x=y+2, abt y-axis

[warwick]

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the y-axis.

23. x = y^2, x = y+2

Here's the integral I worked out.

pi [definite integral of (y+2)^2 - (y^2)^2 with respect to y from -1 to 2]

35. Find the volume of the solid that results when the region enclosed by y = square root of x, y = 0, and x = 9 is revolved about the line x = 9.

Now, I got the correct answer for this one, but I'm a little unclear in explicitly explaining it to myself why I had -9 in my integral.

pi [definite integral of (y^2-9)^2 with respect to y from 0 to 3]

 

[tkhunny]

23. Great! So, what did you get as the result of evaluating that integral?

35. I'd prefer 9 - y^2, but really, what's wrong with "-9" where it's supposed to be. What's the result?

If you really don't like it, do it the other way. Notice how it will make a difference, this time, whether it is 9 - x or x - 9.

\(\displaystyle \L\;2*\pi*\int_{0}^{9}(9-x)\sqrt{x}dx\)

 

[warwick]

23. Great! So, what did you get as the result of evaluating that integral?

35. I'd prefer 9 - y^2, but really, what's wrong with "-9" where it's supposed to be. What's the result?

If you really don't like it, do it the other way. Notice how it will make a difference, this time, whether it is 9 - x or x - 9.

\(\displaystyle \L\;2*\pi*\int_{0}^{9}(9-x)\sqrt{x}dx\)

I got the wrong answer for 23: 123pi/15.

The answer for 35 is 648pi/5. I'm a bit surprised my integral yielded the correct answer. Show me how you got 9 - y^2. Thanks.

 

[tkhunny]

35. \(\displaystyle y = \sqrt{x}\;or\;x = y^{2}\) in the first quadrant.

23. Did you just give up after the first try? Do it again and don't make the same error, whatever it was. Unfortunately, you have not shown your work, so no one can help you find the error. It's just a polynomial. Multiply everything out and do each piece separately. Use the shell method to check your answer. I recommend doing ALL problems both ways. You WILL get better at both in this way.

 

[warwick]

35. \(\displaystyle y = \sqrt{x}\;or\;x = y^{2}\) in the first quadrant.

23. Did you just give up after the first try? Do it again and don't make the same error, whatever it was. Unfortunately, you have not shown your work, so no one can help you find the error. It's just a polynomial. Multiply everything out and do each piece separately. Use the shell method to check your answer. I recommend doing ALL problems both ways. You WILL get better at both in this way.

No. Of course not. I worked through it several times. I only resort to this site after I've exhausted all my ideas (or lack thereof).

Is my starting integral correct for #23?

pi [definite integral of y^2 +4y +4 -y^4 with respect to y from -1 to 2]

For 35, how did you know to include the 9 quantity in the function equation.

 

[tkhunny]

Giving up: Good. Just checking.

2x. Go for it.

3x. Aren't we rotating about x = 9? How far are things from x = 9?

 

[galactus]

Here's a graph of #32:

\(\displaystyle \L\\2{\pi}\int_{0}^{4}[x(\sqrt{x}-x+2)]dx\)

#35: \(\displaystyle \L\\2{\pi}\int_{0}^{9}(x-9)(\sqrt{x}-9)dx\)