## anonymous one year ago Evaluate the integral by interpreting it in terms of areas. S^0 on the bottom -6 (5+root36-x^2)dx I have no idea how to work this problem out, please help!

1. ganeshie8

Take a good look at the integrand, does it look familiar to you ?

2. anonymous

the -6 is on the bottom of the integral, um no not really

3. ganeshie8

like this ?$\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx$

4. anonymous

yes

5. ganeshie8

May be rearrange the integrand a bit $\large y=\color{blue}{5+\sqrt{36-x^2}}$ $\large (y\color{blue}{-5})^2=(\color{blue}{\sqrt{36-x^2}})^2$ $\large \color{blue}{x^2}+(y\color{blue}{-5})^2=\color{blue}{36}$ what about now ? seen this before ?

6. anonymous

it has to do with circles right?

7. ganeshie8

|dw:1436074459700:dw|

8. ganeshie8

look at only the top half, because the original function is $$y=5+\sqrt{36-x^2}$$, which represents only the upper half of circle.

9. anonymous

okay

10. ganeshie8

you need to find the area under that curve between x=-6 and x=0

11. ganeshie8

|dw:1436074742051:dw|

12. ganeshie8

thats the area the given integral represents see if you can work it

13. anonymous

would my answer be 36pi/2? Im not really sure what Im supposed to do, this is the firts problem i have that is like this

14. ganeshie8

how did you get 36pi/2 ?

15. anonymous

1/2pir^2=1/2pi6^2=36pi/2 and then i did it another way and got 30

16. ganeshie8

nope, notice that the area is made up of a "quarter circle" and a "rectangle" : |dw:1436075697415:dw|

17. ganeshie8

find area of quarter circle, find area of rectangle add them up

18. anonymous

okay, what do i do (5+root36-x^2)

19. anonymous

would my answer be 6+9pi? @ganeshie8

20. ganeshie8

whats the area of rectangle ?

21. ganeshie8

|dw:1436076828225:dw|

22. anonymous

is that a 5? so it would be 25

23. ganeshie8

careful, you need to use area of rectangle formula not area of square

24. anonymous

w*l=6*6=36

25. ganeshie8

try again

26. anonymous

30

27. ganeshie8

Yes, save that, next find the area of quarter circle

28. anonymous

1/4PI*6^2=9pi

29. anonymous

@ganeshie8

30. ganeshie8

31. anonymous

30+9pi

32. ganeshie8

$\large \int\limits_{-6}^0 \color{blue}{5+\sqrt{36-x^2}}\, dx~~=6\times 5+\dfrac{\pi\times 6^2}{4} = 30+9\pi$

33. anonymous

Thank you!!!