Question

Sketch the region enclosed by the curves given below. Decide whether to integrate with respect to x or y. Then find the area of the region.
2 y = 5sqrt{x} , y = 3, 2 y + 2 x = 7

Area =

Answer 1

How are you supposed to find the intersections?

Answer 2

These?

\(2y = 5\sqrt{x}\)

y = 3

2y + 2x = 7

Answer 3

yes

Answer 4

Find intersections by taking two at a time and solving.

\(2y=5\sqrt{x}\)
\( y = 3\)

We have y = 3. We just need x. 2(3) = 5sqr(x) ==> sqrt(x) = 6/5 ==> x = 36/25

You do the next pair.

\(2y=5\sqrt{x}\)
\(2y + 2x = 7\)

Answer 5

thats the one i am having trouble for

Answer 6

\[2 y=5 \sqrt{x},4y^2=25x,y^2=\frac{ 25 }{ 4 }x\]
it is a parabola.
\[x=\frac{ 4 }{ 25 }y^2,2y+2*\frac{ 4 }{25 }y^2=7,8y^2+50y=175\]
\[8y^2+50y-175=0,y=\frac{ -50 \pm \sqrt{(50)^2-4*8*-175} }{2*8 }\]
\[y=\frac{ -50\pm \sqrt{2500+5600} }{ 16 }=\frac{ -50\pm \sqrt{8100} }{ 16 }=\frac{ -50\pm 90 }{ 16 }=\frac{ 40 }{ 16 },\frac{ -140 }{16 }\]
\[y=\frac{ 5 }{2 },\frac{ -35 }{4 }\]
\[when y=\frac{ 5 }{ 2 },
as given y is positive

\[required~ area=\int\limits_{5/2}^{3} x~dy\]
\[=\int\limits_{5/2}^{3}\frac{ 7-2y }{ 2 }dy\]
calculate.

Answer 7

oh wow thnks

Answer 8

but i am still troubled with how you got the intersections part.....