## Bladerunner1122 2 years ago The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. h t t p://goo.gl/jXIZD

The functions f and g are given by f(x)=√x and g(x)=6-x. Let R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure in the link below. Please show your work. h t t p://goo.gl/jXIZD 1. Find the area of R. 2. The region R is the base of a solid. For each y, where 0<=y<=2, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 2y. Write, but do not evaluate, an integral expression that gives the volume of the solid. 3. There is a point P on the graph of f at which the line tangent to the graph of f is perpendicular to the graph of g. Find the coordinates of point P.

2. Kamille

3. sirm3d

|dw:1362062051357:dw|

4. Kamille

Well, I will upload a picture

5. Kamille

Painted area you need to find.

6. Kamille

Firstly, you need to find x values, where these two graphis intersects. Do you know how to do it?

7. Kamille

No, how do I do that?

And if possible it should be without the calculator.

11. Kamille

Well, actually, you don't need to, because you have a pont (4;2) given. If you don't have any point, you need to solve this: √x=6-x

12. Kamille

Well, it is easy, you won't need to use a caculator

13. Kamille

Secondly, you need to find there g(x)=6-x intersects Ox line, do you know why you need to do it?

Wouldn't g(x)=6-6?

Since I have the point I don't need to solve for it right?

16. Kamille

no, g(x) intererects Ox when 0=6-x x=6

17. Kamille

|dw:1362062632093:dw|

Mhmm... that make sense so far.

19. Kamille

|dw:1362062739064:dw| So, R consist of two parts: R1 and R2, right?

20. Kamille

If you want to find R=R1+R2, right?

21. Kamille

Area of R1: $A=\int\limits_{0}^{4}\sqrt{x}dx$

22. Kamille

Can you find R1 area for me?

16/3?

24. Kamille

No. Show me how you solve it?

(2x^3/2)/3 + C

26. Kamille

No, $\int\limits_{0}^{4}\sqrt{x}dx=(\frac{ x ^{1,5} }{ 1,5 })|4=\frac{ 4^{1,5}}{ 1,5}$

27. Kamille

Oh, we got the same answer, ahh, sorry! So, now you have area of R1, do you know how to get area of R2?

So now I'm a little confused. Somehow your work doesn't look like my answer? And R2 would be the integral of 6-x dx on the interval [4,6]?

@Kamille

30. sirm3d

|dw:1362064551772:dw| $\Large \int_0^2 (x_R-x_L) dy$ $xR: y=6-x\Rightarrow x=6-y\\xL:y=\sqrt{x}\Rightarrow y^2=x$

For the area of R2 I get 2. So the area for the total should be 16/3 + 2? = 22/3?

32. sirm3d

|dw:1362064821410:dw|$\Large V=\int_0^2A(y) \mathrm dy$