anonymous
  • anonymous
Find the given integral: draw a picture and answer the question based on the picture.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{0}^{4} \sqrt{4-(x-2)^{2}} dx\]
anonymous
  • anonymous
@Hero @freckles @satellite73
anonymous
  • anonymous
@satellite73

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freckles
  • freckles
\[y=\sqrt{4-(x-2)^2} \\ y^2=4-(x-2)^2, y\ge 0 \\ (x-2)^2+y^2=4,y \ge 0 \] so you are only looking at top part of the circle
anonymous
  • anonymous
yes I need to kow the graph! plz help me with this one!
freckles
  • freckles
so you don't know how to draw the circle (x-2)^2+y^2=4?
freckles
  • freckles
this is in center-radius form....
freckles
  • freckles
(x-h)^2+(y-k)^2=r^2 center is (h,k) r is radius and we have y>=0 so erase bottom part of circle because it will fall below the y-axis
anonymous
  • anonymous
|dw:1449721277459:dw|
freckles
  • freckles
right
freckles
  • freckles
area of circle is pi*r^2 and you have half a circle so your area is a half of pi*r^2
freckles
  • freckles
where your r is...
anonymous
  • anonymous
so how do I figure out what my r is?
freckles
  • freckles
r is the radius
anonymous
  • anonymous
yeah but how do I find radius can u explain plz I will need that
anonymous
  • anonymous
I have to show/expalin every step of the procedure
freckles
  • freckles
how were you able to draw the graph above without knowing the radius?
freckles
  • freckles
(x-h)^2+(y-k)^2=r^2 (h,k) is the center and r is the radius...
freckles
  • freckles
your equation is in this exact form
anonymous
  • anonymous
I had the graph but know nothing abotu it but I think we can graph that on the grpahing cal
freckles
  • freckles
example \[(x+3)^2+(y-2)^2=9 \\ \text{ has center } (-3,2) \text{ and radius } 3 \] |dw:1449721603541:dw| ignore the center point it was just to help me to determine the circle
freckles
  • freckles
|dw:1449721667903:dw| any line segment I draw from the center of this circle a point on the circle is going to have length 3 because the radius is 3
anonymous
  • anonymous
alright so the radius in this case is 2
freckles
  • freckles
yes
anonymous
  • anonymous
and now i have to find the area
freckles
  • freckles
just plug into the formula for area of circle and don't forget to take half that amount since you have half the circle
anonymous
  • anonymous
and why is it haalf circle again?
freckles
  • freckles
because we don't have \[y=\pm \sqrt{4-(x-2)^2} \\ \text{ we just have } y=\sqrt{4-(x-2)^2}\]
anonymous
  • anonymous
A=pir^2 A/2=pir^2/2 A=2pi
freckles
  • freckles
so y cannot be negative
freckles
  • freckles
so there should be nothing under the x-axis
freckles
  • freckles
|dw:1449721987418:dw|
freckles
  • freckles
if we had \[y=-\sqrt{4-(x-2)^2}\] then it would have been the bottom half
anonymous
  • anonymous
awesoem I understnd it now! perfect thanks that grph helped me explain it! :) tysm
anonymous
  • anonymous
u are like a god to me today haha was stressed about this question
anonymous
  • anonymous
I am goign to close this question Because have two more questions whichis how to get the local extrema and points of inflection without a calculator

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