Please Help Me Double Check if I have Done These Calculus Probs Correctly??

**nuvydeep** · May 31st 2010, 12:25 AM

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

Please reply if you notice where I have made a mistake... Thanks guys, Much Love

**Prove It** · May 31st 2010, 12:28 AM

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

Please reply if you notice where I have made a mistake... Thanks guys, Much Love

1. should actually be

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

**nuvydeep** · May 31st 2010, 12:33 AM

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??

**Triapod** · May 31st 2010, 02:20 AM

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

$<br /> \pi \int_0^4 {(4 - y} ) dy = 8\pi\<br />$

**HallsofIvy** · May 31st 2010, 03:22 AM

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

$<br /> \pi \int_0^4 {(4 - y} ) dy = 8\pi\<br />$

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: $2\pi\int_0^2 x(4- x^2)dx$ .

**nuvydeep** · May 31st 2010, 08:56 AM

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.

You guys, does anybody else have any comments about my work, any mistakes you notice in the work or answer? Please post..

**nuvydeep** · May 31st 2010, 01:26 PM

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?

**nuvydeep** · May 31st 2010, 08:37 PM

Uh anyone?

**drumist** · June 1st 2010, 04:47 AM

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 $\frac{8}{3}$ .

**drumist** · June 1st 2010, 05:00 AM

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

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

**nuvydeep** · June 1st 2010, 07:22 AM

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

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

Thanks for your reply drumist! At least someone replied..

So from the above ^, I got the area to be 256/15. Is that correct?

**drumist** · June 1st 2010, 07:39 AM

Originally Posted by nuvydeep

So from the above ^, I got the area to be 256/15. Is that correct?

Yes it is

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