• May 31st 2010, 12:25 AM
nuvydeep
Hello my dear lovely fellow math help forum users!!

Can you please check if I have done these Calculus problems correctly or not? I have taken a picture of the problems with my work in writing. You can enlarge the image with this link: http://img62.imageshack.us/img62/323/123niz.jpg

http://img62.imageshack.us/img62/323/123niz.jpg
http://img265.imageshack.us/i/0531000105.jpg/
• May 31st 2010, 12:28 AM
Prove It
Quote:

Originally Posted by nuvydeep
Hello my dear lovely fellow math help forum users!!

Can you please check if I have done these Calculus problems correctly or not? I have taken a picture of the problems with my work in writing. You can enlarge the image with this link: http://img62.imageshack.us/img62/323/123niz.jpg

http://img62.imageshack.us/img62/323/123niz.jpg
http://img265.imageshack.us/i/0531000105.jpg/

1. should actually be

$-\frac{1}{2}\int_0^4{u^{\frac{1}{2}}\,du}$.
• May 31st 2010, 12:33 AM
nuvydeep
Quote:

Originally Posted by Prove It
1. should actually be

$-\frac{1}{2}\int_0^4{u^{\frac{1}{2}}\,du}$.

Oh right.t.. thank you for catching that mistake good mate! See I am thankful for this forum and posting this up :)

Is everything else alright??
• May 31st 2010, 02:20 AM
Triapod
Although i came upon the same answer while doing the third problem, I'm a bit confused as to how you tackled the problem. What is the meaning of the integrand you chose?

I went about it this way:
Solved for x to use the horizontal disks

$
\pi \int_0^4 {(4 - y} ) dy = 8\pi\
$
• May 31st 2010, 03:22 AM
HallsofIvy
Quote:

Originally Posted by Triapod
Although i came upon the same answer while doing the third problem, I'm a bit confused as to how you tackled the problem. What is the meaning of the integrand you chose?

I went about it this way:
Solved for x to use the horizontal disks

$
\pi \int_0^4 {(4 - y} ) dy = 8\pi\
$

He used the "shell" method. Imagine a vertical line at a fixed from (x, 0) to $(x, y)= (x, 4- x^2)$. Rotated around the y- axis, that creates a hollow cylinder. The length of the line is $4- x^2$ and it is rotated in a circle of radius x so circumference $2\pi x$. The surface area of that cylinder is $2\pi x (4- x^2)$. Now if that cylinder has "thickness" dx, the volume of the cylinder is $2\pi x\sqrt(4- x^2)$. Integrating that from 0 to 2 gives the volume: $2\pi\int_0^2 x(4- x^2)dx$.
• May 31st 2010, 08:56 AM
nuvydeep
Quote:

Originally Posted by HallsofIvy
He used the "shell" method. Imagine a vertical line at a fixed from (x, 0) to $(x, y)= (x, 4- x^2)$. Rotated around the y- axis, that creates a hollow cylinder. The length of the line is $4- x^2$ and it is rotated in a circle of radius x so circumference $2\pi x$. The surface area of that cylinder is $2\pi x (4- x^2)$. Now if that cylinder has "thickness" dx, the volume of the cylinder is $2\pi x\sqrt(4- x^2)$. Integrating that from 0 to 2 gives the volume: $2\pi\int_0^2 x(4- x^2)dx$.

Yes that's right I used the shell method.

• May 31st 2010, 01:26 PM
nuvydeep
I feel like I got the first one wrong... shouldn't the domain be switch from 4 to 0 to 0 to 4 and flip the whole integrand?
• May 31st 2010, 08:37 PM
nuvydeep
Uh anyone?
• Jun 1st 2010, 04:47 AM
drumist
nuvydeep, your original work for the first problem was correct. Since you had reversed the order of the endpoints, you did not need the negative sign in the front.

However, in your haste, you may have written the final answer wrong. The final expression should be $\frac{1}{3} 4^{3/2}$. Notice that $4^{3/2} = (2^2)^{3/2} = 2^3 = 8$, so the final answer is simply $\frac{8}{3}$.
• Jun 1st 2010, 05:00 AM
drumist
For the second problem, the cross sections appear to be 45-45-90 right triangles, meaning the height and base are equal, which should give us an area of

$A=\frac{1}{2} (4-x^2)^2$

So, the integral would be

$\frac{1}{2} \int_{-2}^2 (4-x^2)^2 \, dx = \int_0^2 (16-8x^2+x^4) \, dx$
• Jun 1st 2010, 07:22 AM
nuvydeep
Quote:

Originally Posted by drumist
For the second problem, the cross sections appear to be 45-45-90 right triangles, meaning the height and base are equal, which should give us an area of

$A=\frac{1}{2} (4-x^2)^2$

So, the integral would be

$\frac{1}{2} \int_{-2}^2 (4-x^2)^2 \, dx = \int_0^2 (16-8x^2+x^4) \, dx$