# Thread: Integration Application: Volume generated by area bounded by two curves

1. ## Integration Application: Volume generated by area bounded by two curves

Q: Find the volume generated by area bounded by two curves rotated through 360 degrees about the y-axis:
a) y = x²
y = 2- x²

b) y = x2
y2 = 8x

For these two parts, which formulae should I use?

Volume generated

= π ∫ x² dy
or = π ∫ [f(y)]² – [g(y)]² dy ?

For part a, it will be simplier if I use the first formula where I simplify the y before using the first formula:

y = 2 - x² - x²
y = 2 - 2x²
x = √ (2 - y)/2

After using the first formula, my answer is π.

However, if I use the second formula, I got diferent answer:

y = x²
x = √y

y = 2 - x²
x = √(2 - y)

V = π ∫ [f(y)]² – [g(y)]² dy
= π ∫ [√y]² - [√(2 - y)]² dy
= .....
= π (4 - 4)
= 0

For part b, it seems that I can't use the first formula, as after simplify,
y = √(8x) - x²
it's hard to change to make x the subject in terms of y.

Using the second formula for part b, I got the answer 24/7π.

So how do we know which formula to be used for which question? Because the question is involving volume generated by area bounded by two curves, shouldn't we be using the second formula only? Why using both formulas I get different answers? Please help me check if my answer is correct.

Many thanks in advance!! =)

2. For problem a)

${y_1} = {x^2},{\text{ }}{y_2} = 2 - {x^2}.$

${y_1} = {y_2} \Leftrightarrow {x^2} = 2 - {x^2} \Leftrightarrow {x^2} = 1 \Leftrightarrow x = \pm 1.$

Shell method

${V_y} = 2\pi \int\limits_0^1 {x \cdot {y_2}dx} - 2\pi \int\limits_0^1 {x \cdot {y_1}dx} = 2\pi \int\limits_0^1 {x\left( {{y_2} - {y_1}} \right)dx} =$

$= 2\pi \int\limits_0^1 {x\left( {2 - 2{x^2}} \right)dx} = 4\pi \int\limits_0^1 {\left( {x - {x^3}} \right)dx} =$

$= \left. {4\pi \left( {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{4}} \right)} \right|_0^1 = \left. {\pi \left( {2{x^2} - {x^4}} \right)} \right|_0^1 = \pi {\text{ (cubic units)}}{\text{.}}$

Look this picture

3. For problem b)

${y_1} = {x^2},{\text{ }}y_2^2 = 8x.$

$y_2^2 = 8x \Leftrightarrow {y_2} = \pm \sqrt {8x} .$

${y_1} = {y_2} \Leftrightarrow {x^2} = \sqrt {8x} \Leftrightarrow \left[ \begin{gathered}{x_1} = 0, \hfill \\{x_2} = 2. \hfill \\ \end{gathered} \right.$

Shell method

${V_y} = 2\pi \int\limits_0^2 {x\left( {\sqrt {8x} - {x^2}} \right)dx} = 2\pi \int\limits_0^2 {\left( {2\sqrt 2 \sqrt {{x^3}} - {x^3}} \right)dx} =$

$= \left. {2\pi \left( {2\sqrt 2 \cdot \frac{{\sqrt {{x^5}} }}{{{5 \mathord{\left/{\vphantom {5 2}} \right.\kern-\nulldelimiterspace} 2}}} - \frac{{{x^4}}}{4}} \right)} \right|_0^2 = \left. {\pi \left( {\frac{{8\sqrt 2 }}{5}\sqrt {{x^5}} - \frac{{{x^4}}}{2}} \right)} \right|_0^2 =$

$= \left( {\frac{{8\sqrt 2 }}{5}\sqrt {32} - 8} \right)\pi = \left( {\frac{{64}}{5} - 8} \right)\pi = \frac{{24}}{5}\pi {\text{ (cubic units)}}{\text{.}}$

4. I see. Thank you =)! I just learnt the shell method from you. But for part a, why did u choose 0 and 1 as the limits (a and b) instead of -1 and 1?

But what if I need to use the disk method? Which formula should I be using and why the answer differs if I use both the methods I mentioned?

5. Besides, I tried the Shell method on this question but I got stuck...

Q: Find the volume generated when region bounded by y=2x² + 5 and
y=3x² + 1 rotated through 2 right angles about the x-axis.

For rotation about the x-axis, we have to change x the subject in terms of y right? Then I got 2π∫ y. √[(y-5)/2] - √[(y-1)/3] dy and have no idea how to proceed...

6. Originally Posted by Cathelyn13
I see. Thank you =)! I just learnt the shell method from you. But for part a, why did u choose 0 and 1 as the limits (a and b) instead of -1 and 1?
Because
...curves rotated through 360 degrees about the y-axis

7. Oh I see. Thanks!! =)

8. If I am to use the Shell method to solve this question, how?

Q: Find the volume generated when region bounded by y=2x² + 5 and
y=3x² + 1 rotated through 2 right angles about the x-axis.

For rotation about the x-axis, we have to change x the subject in terms of y right? Then I got 2π∫ y. √[(y-5)/2] - √[(y-1)/3] dy and have no idea how to proceed...