# Calc 2 Problems

• April 7th 2008, 01:41 PM
Ideasman
Calc 2 Problems
1.) Set up the integral to find length of curve given $y = \sqrt{x}, x =0, x=4$

2.) Set up the integral to find volume if $\int_{0}^{4} \sqrt{x}\, dx$ is rotated around the x-axis.

2.) Set up the integral to find volume if $\int_{0}^{4} \sqrt{x}\, dx$ is rotated around the y-axis.
• April 7th 2008, 01:50 PM
Mathstud28
Quote:

Originally Posted by Ideasman
1.) Set up the integral to find length of curve given $y = \sqrt{x}, x =0, x=4$

2.) Set up the integral to find volume if $\int_{0}^{4} \sqrt{x}\, dx$ is rotated around the x-axis.

2.) Set up the integral to find volume if $\int_{0}^{4} \sqrt{x}\, dx$ is rotated around the y-axis.

1) you are trying to find arclength so you appy the formula $\int_0^4{\sqrt{1+\bigg(\frac{1}{2\sqrt{x}}\bigg)^2 }}dx$

for 2. I assume this is what you mean...from 0 to 4 with $\sqrt{x}$...so for x-axis we apply the formula to get ${\pi}\int_0^4{x}dx$

and for 3. we apply the shell method formula to attain ${2\pi}\int_0^4{x\cdot{\sqrt{x}}}dx$
• April 7th 2008, 02:04 PM
Ideasman
Quote:

Originally Posted by Mathstud28
1) you are trying to find arclength so you appy the formula $\int_0^4{\sqrt{1+\bigg(\frac{1}{2\sqrt{x}}\bigg)^2 }}dx$

for 2. I assume this is what you mean...from 0 to 4 with $\sqrt{x}$...so for x-axis we apply the formula to get ${\pi}\int_0^4{x}dx$

and for 3. we apply the shell method formula to attain ${2\pi}\int_0^4{x\cdot{\sqrt{x}}}dx$

Sorry mistyped the question. Originally, the question was:

1.) Compute the area bounded by $y=\sqrt{x}, y= 0$ and $x=4$.

Then, for 2 and 3 it was

2.) Set up the integral to find the volume if this AREA is rotated about the x-axis

3.) Set up the integral to find the volume if this AREA is rotated about the y-axis

This answer still apply? I think 1 will be different, as what you gave me for #1 will be the length of the curve (arclength)

Thanks a lot
• April 7th 2008, 02:15 PM
Mathstud28
Quote:

Originally Posted by Ideasman
Sorry mistyped the question. Originally, the question was:

1.) Compute the area bounded by $y=\sqrt{x}, y= 0$ and $x=4$.

Then, for 2 and 3 it was

2.) Set up the integral to find the volume if this AREA is rotated about the x-axis

3.) Set up the integral to find the volume if this AREA is rotated about the y-axis

This answer still apply? I think 1 will be different, as what you gave me for #1 will be the length of the curve (arclength)

Thanks a lot

THe only one that changes is the first one it just becomes $\int_0^4{\sqrt{x}}dx$
• April 7th 2008, 02:20 PM
Ideasman
Quote:

Originally Posted by Mathstud28
for 2. I assume this is what you mean...from 0 to 4 with $\sqrt{x}$...so for x-axis we apply the formula to get ${\pi}\int_0^4{x}dx$

Thanks a lot. Final question, for this, did you mean $\sqrt{x}$ or simply $x$?

Edit: Oh I see why its just x.