Calculus II Volumes of Revolution and Basic Integration Questions

[ardentmed]

Hey guys,

I have a couple of questions about the problem set I'm doing at the moment. Although I was able to solve most of these, I'm doubting quite a few of my responses.

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For the first one, I split up the integral into two parts and obtained three different functions from the three coordinates, then added the area obtained from the two integrals, 4 and 6 respectively, to get 10.

For the second one, I used the formula for a semi-circles area, $\pi * r ^2 / 2 and obtained 9/8 * $\pi for the area, then integrated it from y=2 to y=0 with respect to y and obtained 9/4 * $\pi in total. This is because the triangle has a base of 3 metres in length, so I presumed that I could use that (and divide it by 2) to get the radius of the semi-circle. Albeit, I'm really not too sure about this one

For the third one, I just used the cylindrical shells method with (2+x) as the radius, expanded the integral, and integrated. I got ~268.083 as the answer.

For the fourth one, I got 29 $\pi / 30 for both answers, so I'm pretty sure I got it right.

For number eight, I integrated 245,000 $\pi * x from 0.3m to 1.5 metres since that is how far the water must travel to get out of the tub. I ended up getting 264,000 Joules, albeit I feel as if I may have made an error somewhere along the way.

Number nine should have been relatively easy, and is a simple average value question, so it's just 1/(b-a) * \int_{}^{} \,d , which gave me 24 degrees celcius.

Thanks in advance for the help, guys. I really appreciate it

 

[ineedhelpnow]

Just want to point out when ur using symbols or commands you have to put a dollar sign at the beginning and at the end. For ex. Dollar sign \pi dollar sign gives you $\pi$ kind of easy to understand the post that way.

 

[ineedhelpnow]

Lol type your command and then end it. Say what you want to say in words and start again. so, dollar sign \pi dollar sign and then bla bla bla dollar sign \theta dollar sign otherwise it'll put everything in latex format

 

[ineedhelpnow]

For number 9 $\int_{0}^{12} \ 20+6 \sin\left({\frac{\pi t}{12}}\right)dt$ and then divide you answer by 12 to get the avg. My final answer is 258.837/12=23.82

 

[MarkFL]

Hey guys,

I have a couple of questions...

Hello and welcome to MHB! :D

In the future, please post no more than two questions at the beginning of a thread. Threads are easier to follow this way, without becoming convoluted if more than one person is trying to help with different questions at the same time, etc.

For the first one, we can check your answer using a formula from coordinate geometry for the area of a triangle given 3 non-collinear points:

\(\displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|\)

\(\displaystyle A=\frac{1}{2}\left|(-2-0)(3-4)-(3-0)(-2-4) \right|=\frac{1}{2}\left|2+18 \right|=10\)

Correct! :D

For the second problem, the hypotenuse of the triangular base lies along the line:

\(\displaystyle y=-\frac{3}{2}x+2\)

Solving this for the radii of the slices, which is \(\displaystyle \frac{x}{2}\), we find:

\(\displaystyle \frac{x}{2}=\frac{2-y}{3}\)

Now, the volume of an arbitrary slice is:

\(\displaystyle dV=\pi\left(\frac{2-y}{3}\right)^2\,dy\)

and adding up all the slices, we obtain:

\(\displaystyle V=\frac{\pi}{9}\int_0^2(2-y)^2\,dy\)

Let:

\(\displaystyle u=2-y\,\therefore\,du=-dy\)

and we have:

\(\displaystyle V=\frac{\pi}{9}\int_0^2 u^2\,du\)

What do you get when you evaluate this integral?

 

[ardentmed]

Hello and welcome to MHB! :D

In the future, please post no more than two questions at the beginning of a thread. Threads are easier to follow this way, without becoming convoluted if more than one person is trying to help with different questions at the same time, etc.

For the first one, we can check your answer using a formula from coordinate geometry for the area of a triangle given 3 non-collinear points:

\(\displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|\)

\(\displaystyle A=\frac{1}{2}\left|(-2-0)(3-4)-(3-0)(-2-4) \right|=\frac{1}{2}\left|2+18 \right|=10\)

Correct! :D

For the second problem, the hypotenuse of the triangular base lies along the line:

\(\displaystyle y=-\frac{3}{2}x+2\)

Solving this for the radii of the slices, which is \(\displaystyle \frac{x}{2}\), we find:

\(\displaystyle \frac{x}{2}=\frac{2-y}{3}\)

Now, the volume of an arbitrary slice is:

\(\displaystyle dV=\pi\left(\frac{2-y}{3}\right)^2\,dy\)

and adding up all the slices, we obtain:

\(\displaystyle V=\frac{\pi}{9}\int_0^2(2-y)^2\,dy\)

Let:

\(\displaystyle u=2-y\,\therefore\,du=-dy\)

and we have:

\(\displaystyle V=\frac{\pi}{9}\int_0^2 u^2\,du\)

What do you get when you evaluate this integral?

8pi / 27. Thanks a ton!

 

[MarkFL]

8pi / 27. Thanks a ton!

Correct! (Yes)

Let's look at the third one...you gave your answer and $r$, but I don't really know what you did, so let's step through it and check your result.

I would begin by plotting the region to be revolved and the axis of rotation:

View attachment 2770

Now, the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

Can you state $h$ as a function of $x$ and then state the volume of the arbitrary shell as a function of $x$ using $h(x)$ and $r(x)$? After that sum up the shells in a definite integral...what do you have up to that point?

 

[ardentmed]

Correct! (Yes)

Let's look at the third one...you gave your answer and $r$, but I don't really know what you did, so let's step through it and check your result.

I would begin by plotting the region to be revolved and the axis of rotation:

View attachment 2770

Now, the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

Can you state $h$ as a function of $x$ and then state the volume of the arbitrary shell as a function of $x$ using $h(x)$ and $r(x)$? After that sum up the shells in a definite integral...what do you have up to that point?

So, to reiterate my response to the question, the radius would be 2+x, so that replaces the radius in the formula. Also, the greater function is subtracted by the function that is smaller over that interval in which they intersect. Ultimately, after integrating, wouldn't that end up giving me ~268.1?

 

[MarkFL]

So, to reiterate my response to the question, the radius would be 2+x, so that replaces the radius in the formula. Also, the greater function is subtracted by the function that is smaller over that interval in which they intersect. Ultimately, after integrating, wouldn't that end up giving me ~268.1?

Well, your answer is correct to 1 decimal place, but can you show the integral you used, your work and the exact answer? :D