Compute the integral ∫C(O,2)2xydx+(2x+3y2+2y3)dy, where C(O,2) is the circle of center O and radius 2, with the positive orientation.

depends on how many days you are talking about

step-by-step explanation:

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

what are ur choices?

the answer is:

the lateral area is equal to

to find the lateral area of the square pyramid, first we need to find the perimeter of the base, and then, find the lateral area using the following formulas:

where,

s, is the base length

slant height, is the height of a the side of the pyramid

we are given:

now, calculating the perimeter of the base, we have:

then, calculating the lateral area:

hence, we have that the lateral area is equal to

have a nice day!