Thread: Integrating enclosed area

1. Integrating enclosed area

Calculate the area enclosed by x = 9-y^2 and x = 5 in two ways: as an integral along the y-axis and as an integral along the x-axis.

Please tell me if my figure is correct, and also if the shaded area is also the are I am looking for.
Also is there any formulas I need to use for this problem ? How do I do this both ways ?
Thank You

2. Yes, your graph is fine.

To integrate over the x-axis, you need

$\displaystyle\int_{5}^9f(x)dx$

so $x=9-y^2\Rightarrow\ y^2=9-x\Rightarrow\ y=\pm\sqrt{9-x}=f(x)$

Take the positive one and double your result as the x-axis is an axis of symmetry.

$2\displaystyle\int_{5}^9\left(9-x\right)^{0.5}dx$

To integrate over the y-axis, you could integrate $f(y)$ from $y=-3$ to $y=3.$

If you like, again use the x-axis as an axis of symmetry and double the integral from $y=0$ to $y=3.$

This integral includes the unshaded part against the y-axis, so you have a few ways to cope with that.

The easiest way is to subtract 5 from $x=f(y)$

$x-5=f(y)-5=4-y^2$

The new function is $x=4-y^2$

so the limits of integration become $\pm2$

You can then calculate

$2\displaystyle\int_{0}^2\left(4-y^2\right)}dy$

3. Thank You very much. I'll give it a shot.

4. The first integral can be evaluated by making a substitution,
while the 2nd one doesn't require any substitution.

5. Thank you for all you help.

6. Ok so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?

7. Originally Posted by wair
Ok so I solved the first one I get .5. I let u = 9-x , du = -dx so I multiplied by a negative inside the integral and a negative outside. then I get, -2(.5(9-x)^-.5). Then using the fundemental theorem of calculus I used the limts 5 to 9 and my final answer was .5. is this correct ?
Not quite,
you differentiate to introduce the substitution,
but you are also differentiating when you should be integrating!

$u=9-x\Rightarrow\ du=-dx\rightarrow\ dx=-du$

$x=5\Rightarrow\ u=4$

$x=9\Rightarrow\ u=0$

the integral becomes

$\displaystyle\ -2\int_{4}^0u^{0.5}}du=(-2)\left[-\int_{0}^4u^{0.5}}du\right]=2\int_{0}^4u^{0.5}}du$

$=2\displaystyle\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]$ from u=0 to u=4

8. so I just had to chnage my limits ? and then invert them to get rid of the what was it called "signed area" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?

9. Originally Posted by wair
so I just had to chnage my limits ? and then invert them to get rid of the what was it called "signed area" ? But I don't understand how you got the last step. wouldn't it be .5(u)^-.5 ?
No, that's what you get when you differentiate.

$\displaystyle\frac{d}{du}u^{0.5}=0.5u^{-0.5}$

But

$\displaystyle\int{u^{0.5}}du=\frac{u^{1.5}}{1.5}+c$

because

$\displaystyle\frac{d}{du}\left[\frac{u^{1.5}}{1.5}+c\right]=\frac{1.5}{1.5}u^{0.5}$

To integrate, apply differentiation in reverse.

10. Oh right right. I understand now thank you .

11. Ok so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?

12. I get 5.3 for the second one. shouldn't I get the same answer for both ?

13. Originally Posted by wair
Ok so for the first one I get -21.1. (1/1.5)(9-x)^1.5. 2(7.45)-2(18)=-21.1. Is that correct ?
If you work with 9-x instead of u, then there is no need to change the limits of integration.

$2\displaystyle\int_{x=5}^{x=9}{(9-x)^{0.5}}dx=-2\int_{x=5}^{x=9}{(9-x)^{0.5}}d(9-x)$

$=2\displaystyle\int_{x=9}^5{(9-x)^{0.5}}d(9-x)=2\frac{(9-x)^{1.5}}{1.5}$ from x=9 to 5 (start calculations at x=5)

$=2\displaystyle\frac{(9-5)^{1.5}-(5-5)^{1.5}}{1.5}=2\frac{4^{1.5}}{1.5}=2\frac{8}{1.5} =2\frac{16}{3}=\frac{32}{3}$

You must get a positive value for area, so as our graph is symmetrical about the x-axis, we integrate the part above the x-axis and double it.

Using the u-substitution, we get

$\displaystyle\ 2\int_{0}^4{u^{0.5}}du=2\left[\frac{u^{1.5}}{1.5}\right]$ from u=0 to u=4 (start calculations at u=4)

$\displaystyle\ =2\frac{4^{1.5}}{1.5}-0=2\frac{8}{1.5}=\frac{16}{1.5}=\frac{32}{3}$

For the 2nd integral..

$\displaystyle\ 2\int_{0}^2{\left(4-y^2\right)}dy=2\int_{0}^2{4}dy-2\int_{0}^2{y^2}dy=2\left[4y-\frac{y^3}{3}\right]$ evaluated from y=0 to y=2

(start calculations at y=2)

comes out as the same value.
Review your calculations for both integrals.

14. Thank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you

15. Originally Posted by wair
Thank you very much. So if I want to use the new limits I don't need to replacex u with what u equals. And both values must be the same. OK Thank you
I think you mean.... "don't need to replace u with 9-x".

Yes, it is simplest to work with "u", having made the substitution, then integrate f(u)du using the "u" limits.

This area you are now calculating is on a different graph but evaluating the new "u" integral will give the same area as the shaded region on the original graph.

If you prefer not to change the limits, here's how to do it....

$\displaystyle\ -2\int_{x=5}^{x=9}{u^{\frac{1}{2}}}du=-2\left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]$ from x=5 to x=9 (start calculations at x=9)

$=-\displaystyle\frac{4}{3} (9-x)^{\frac{3}{2}}$ from x=5 to x=9

$=-\displaystyle\frac{4}{3}\left[ (9-9)^{1.5}-(9-5)^{1.5}\right]=\frac{32}{3}$

You've got to practice. Keep going until you can reproduce the solution for both integrals.