problems with the y axis intergration (1 Viewer)

[red802]

1) calculate the area bounded by the graph of y=1/4x^2, the axis and the lines y = 0, y=1 lying in the first quadrant ( what is the first quadrant)

2) calculate the area enclosed by the curve y=4x^2, the y-axis and the lines y=1, y=4, lying in the first quadrant

3) find the area between y=1/x^2, the y axis and the lones y=1, y=2. LEave the answer in surd form

if u can answer tthese, can u show the working out

 

[insert-username]

For all of them, you need to rewrite the equation in terms of y, i.e:

1) y=1/4x², therefore x = 1/2√y. Now, it's exactly the same as integrating over the x-axis - F(b) - F(a). Integrate from y = 0 to y = 1 (the first quadrant is the top right hand side of your cartesian plane, up and right from the origin).

2) y=4x², therefore x = √y/2. Integrate from y=1 to y=4.

3) y=1/x², therefore x = 1/√y. Integrate from y=1 to y=2.


I_F

 

[red802]

sry about this, i still got problems trying to solve the equations, can u give us a little bit more help

 

[Mountain.Dew]

sry about this, i still got problems trying to solve the equations, can u give us a little bit more help

u need to rewrite the equation so that u have ur function in terms of y's, not x's. so, u need to do a bit of algebraic rearranging. get x as the subject

so for 1, we have y=1/4x² --> first times both sides by 4

so, we get 4y = x², then we square root both sides to get the equation with x as the subject ==> x = 2y^1/2

so, u have ur area = integral from y=0 to y=1 OF x = 2y^1/2 dx

the 1st quadrant refers to the top right hand "quarter" of the xy plane, also known as the cartesian plane. 2nd quadrant is top left hand side, 3rd is bottom left hand side, 4th is bottom right hand side. SO, numbering 1 --> 4 goes anti-clockwise around the cartesian plane.

 

[red802]

thanks guys