#### 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

#### Prove It

MHF Helper
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

1. should actually be

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

nuvydeep

#### nuvydeep

1. should actually be

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

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

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

#### HallsofIvy

MHF Helper
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

$$\displaystyle \pi \int_0^4 {(4 - y} ) dy = 8\pi\$$
He used the "shell" method. Imagine a vertical line at a fixed from (x, 0) to $$\displaystyle (x, y)= (x, 4- x^2)$$. Rotated around the y- axis, that creates a hollow cylinder. The length of the line is $$\displaystyle 4- x^2$$ and it is rotated in a circle of radius x so circumference $$\displaystyle 2\pi x$$. The surface area of that cylinder is $$\displaystyle 2\pi x (4- x^2)$$. Now if that cylinder has "thickness" dx, the volume of the cylinder is $$\displaystyle 2\pi x\sqrt(4- x^2)$$. Integrating that from 0 to 2 gives the volume: $$\displaystyle 2\pi\int_0^2 x(4- x^2)dx$$.

#### nuvydeep

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

Yes that's right I used the shell method.

#### 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?

Uh anyone?

#### 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 $$\displaystyle \frac{1}{3} 4^{3/2}$$. Notice that $$\displaystyle 4^{3/2} = (2^2)^{3/2} = 2^3 = 8$$, so the final answer is simply $$\displaystyle \frac{8}{3}$$.

nuvydeep

#### 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

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

So, the integral would be

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