# Thread: Area Using Integrals, Disk Method, and Shell Method

1. ## Area Using Integrals, Disk Method, and Shell Method

Hello everyone, I am new to this forum. That being said, I am sure I'm about to ask for way too much. I have four calculus problems that I have completed, but I am not sure that I got the right answers. If anyway possible, I would be very appreciative if someone looked at my answers, and told me if they were correct, and if not, showed me how to do the problem correctly.

Two problems have to do with area of a 2-d region, and two problems have to do with volume of a rotated area.

For Problems 1 & 2: Write the integral, or integrals, for the area, and evaluate your integral(s).

Problem 1. R is the region bounded by y = $x^2$ +4, y = -2x, x = 0, and x = 1.

My Solution

$R_{1}$ is the region bounded by x = 1 and x = $\frac{-1}{2}$y from - 2 to 0 on the y-axis.

A( $R_{1}$) = $\int^0_{-2}$(1 + $\frac{1}{2}y$) dy = [y + $\frac{y^2}{4}$ $]^0_{-2}$ = 0 + 1 = 1

$R_{2}$ is the region bounded by x = 1 and x = 0 from 0 to 4 on the y-axis.

A( $R_{2}$) = $\int^4_0$(1) dy = [y $]^4_0$ = 4 - 0 = 4

$R_{3}$ is the region bounded by x = 1 and x = $\sqrt{y - 4}$ from 4 to 5 on the y-axis.

A( $R_{3}$) = $\int^5_4$(1 - $\sqrt{y - 4}$) dy = [y - $\frac{2(y - 4)^{\frac{3}{2}}}{3}$ $]^5_4$ = (5 - $\frac{2}{3}$) - (4) = $\frac{1}{3}$

The area of R is A( $R_{1}$) + A( $R_{2}$) + A( $R_{3}$)

1 + 4 + $\frac{1}{3}$ = $\frac{16}{3}$

Problem 2. R is the region bounded by $y^2$ = x + 1, and the line x + y - 5 = 0.

My Solution

$R_{1}$ is the region bounded by y = $\sqrt{x + 1}$ and y = 0 from -1 to 3 on the x-axis.

A( $R_{1}$) = $\int^3_{-1}$ $\sqrt{x + 1}$ dx = [ $\frac{2(x + 1)^{\frac{3}{2}}}{3}$ $]^3_{-1}$ = $\frac{16}{3}$ - 0 = $\frac{16}{3}$

$R_{2}$ is the region bounded by y = - $\sqrt{x + 1}$ and y = 0 from -1 to 3 on the x-axis.

A( $R_{2}$) = $\int^3_{-1}$ $\sqrt{x + 1}$ dx = [ $\frac{2(x + 1)^{\frac{3}{2}}}{3}$ $]^3_{-1}$ = $\frac{16}{3}$ - 0 = $\frac{16}{3}$

$R_{3}$ is the region bounded by x + y - 5 = 0 and y = 0 from 3 to 5 on the x-axis.

A( $R_{3}$) = $\int^5_3$(-x + 5) dx = [ $\frac{-x^2}{2}$ + 5x $]^5_3$ = ( $\frac{-25}{2}$ + 25) - ( $\frac{-9}{2}$ + 15) = 2

$R_{4}$ is the region bounded by y = 0 and y = - $\sqrt{x + 1}$ from 3 to 8 on the x-axis.

A( $R_{4}$) = $\int^8_3$ $\sqrt{x + 1}$ dx = [ $\frac{2(x + 1)^{\frac{3}{2}}}{3}$ $]^8_3$ = 18 - $\frac{16}{3}$ = $\frac{38}{3}$

The area of R is A( $R_{1}$) + A( $R_{2}$) + A( $R_{3}$) + A( $R_{4}$)

$\frac{16}{3}$ + $\frac{16}{3}$ + 2 + $\frac{38}{3}$ = $\frac{76}{3}$

For Problems 3 & 4: R is the region bounded by y = $x^2$ + 2, y = x, x = 0, and x = 2. Use the indicated method to write an integral, or integrals, for the volume of the solid of revolution if R is revolved in the indicated way. Then calculate the volume using either method (a), or method (b).

*** I think I got these wrong, so any help whatsoever would be much appreciated ***

Problem 3. Revolve about the x-axis. a. Disk Method b. Shell Method

My Solution

a. Disk Method

V = $\pi$ $\int^2_0$( $(x^2 + 2)^2$ - $x^2$) dx

b. Shell Method (Solved with Shell Method)

V = 2 $\pi$ $\int^2_0$(2 $x^2$ + 4) dx = 2 $\pi$[ $\frac{2x^3}{3}$ + 4x $]^2_0$ = 2 $\pi$( $\frac{16}{3}$ + 8) = $\frac{80\pi}{3}$

Problem 4. Revolve about the y-axis. a. Disk Method b. Shell Method

My Solution

a. Disk Method

V = $\pi$ $\int^2_0$ $(y)^2$ dy + $\pi$ $\int^6_2$ $(2 - \sqrt{2 - y})^2$ dy = $\pi$ $\int^2_0$ $y^2$ dy + $\pi$ $\int^6_2$ $(2 - \sqrt{2 - y})^2$ dy

b. Shell Method (Solved with Shell Method)

V = 2 $\pi$ $\int^2_0$2y dy + 2 $\pi$ $\int^6_2$8 dy = 2 $\pi$[ $y^2$ $]^2_0$ + 2 $\pi$[8y $]^6_2$= 2 $\pi$(4) + 2 $\pi$(48 - 16) = 8 $\pi$ + 64 $\pi$ = 72 $\pi$

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Once again, any help would be greatly appreciate. Thank you very much in advance.

2. For the first two problems, your methodology is mostly correct, but you are definitely taking the hard route. For example, on problem 1, it's possible to define it as a single region if you integrate in the x-direction, like so:

$\int_0^1 \left[ (x^2+4)-(-2x) \right]dx = 16/3$

Your solution was correct, but much more work.

The second problem is wrong though. Again, it can be done with a single region, this time in the y-direction:

$\int_{-3}^2 \left[ (-y+5)-(y^2-1) \right]dy = 125/6$

For the third problem, the disk method is correct. For the shell method, you are integrating on the wrong variable!

$2\pi \int_0^2 y^2 \; dy + 2\pi \int_2^6 y \left( 2-\sqrt{y-2} \right) dy$

For the last problem, the disk method is ALMOST correct. Here's what it should look like:

$\pi \int_0^2 y^2 \; dy + \pi \int_2^6 \left[ 2^2 - \left( \sqrt{y-2} \right)^2 \right]dy$

And finally, you again did the shell method with the wrong variable:

$2\pi \int_0^2 x \left[ (x^2+2)-x \right] dx$

If you have questions about how some of these were set up, please ask. Notice that if the shell method integrates on x, then the disk method will integrate on y, and vice versa. In particular focus on how I set up the shell method and see if it makes sense to you.

3. Wow, I can't believe someone replied, let alone corrected my work. I greatly appreciate it.

On the third problem, aren't the integral bounds supposed to be $\int^6_2$ instead of $\int^4_2$? And on the fourth problem, isn't it supposed to be $x^2$ + 2 instead of $x^2$ - 2?

Also, on the third problem, I get different answers depending on what method I use. With the disk method, I keep getting $\frac{112\pi}{5}$ and with the shell method I get $\frac{32\pi}{3}$

On the fourth problem I get $\frac{32\pi}{3}$ so I'm assuming that my shell method answer for number three is the correct answer since it makes sense that they would both have the same value. I don't know what I'm doing wrong with the disk method though.

4. I take that back, it shouldn't make sense that they would have the same volume. As I'm pretty sure I'm calculating the volume with the disk method correctly, I still can't figure out why I'm getting the wrong answer with the shell method.

5. Originally Posted by mturner07
On the third problem, aren't the integral bounds supposed to be $\int^6_2$ instead of $\int^4_2$? And on the fourth problem, isn't it supposed to be $x^2$ + 2 instead of $x^2$ - 2?
Yes, you're correct on both of these. I just made a few typos while writing it up.

Also, on the third problem, I get different answers depending on what method I use. With the disk method, I keep getting $\frac{112\pi}{5}$ and with the shell method I get $\frac{32\pi}{3}$
The correct answer on problem 3 is $\frac{112\pi}{5}$, so let's look at the shell method calculation:

$2\pi \left[ \int_0^2 y^2 \; dy + \int_2^6 y \left( 2-\sqrt{y-2} \right) dy \right]$

$=~ 2\pi \left[ \int_0^2 y^2 \; dy + \int_2^6 2y \; dy - \int_2^6 y \sqrt{y-2} \; dy \right]$

$=~ 2\pi \left[ \int_0^2 y^2 \; dy + \int_2^6 2y \; dy - \int_0^4 (u+2) \sqrt{u} \; du \right]$

$=~ 2\pi \left[ \int_0^2 y^2 \; dy + \int_2^6 2y \; dy - \int_0^4 u^{3/2} \; du - \int_0^4 2u^{1/2} \; du \right]$

Can you finish it from here?

On the fourth problem I get $\frac{32\pi}{3}$ so I'm assuming that my shell method answer for number three is the correct answer since it makes sense that they would both have the same value. I don't know what I'm doing wrong with the disk method though.
The two solutions for number 4 should be the same, but not the same as the solution from number 3. In each problem you are revolving around a different axis which results in a totally different resulting shape.

The solution for the two parts of problem 4 should give you $\frac{32\pi}{3}$ so that looks OK.

6. Yes. Thank you, I really appreciate your help. I'll be back on these forums often :P