integration

[red and white kop!]

find the area in the first quadrant bounded by the x and y axes, the curve y=x^2 and the line y=16
A using vertical elements
B using horizontal elements and the equation of the curve in the form
x=sqrt y

in the first case i found 21 1/3 and in te second case 42 2/3, i.e. the double
why is this? apparently both answers are 42 2/3, bu i just cannot find that using in A!
im supposed to use integration btw, so in A i calculate the integral of x^2 wrt x within the limits 4 to 0. why is this incorrect?

 

[mmm4444bot]

You're not finding the area of the correct region, in part (A), because your integrand is wrong.

Draw a picture!

What you've done is to find the area of the region bounded by the x-axis, the curve y = x^2, and the line x = 4.

In fact, if you draw a picture, you'll see a rectangular region that's 4 units by 16 units. It's area is 64 square units.

Well, lo-and-behold. If we subtract your result of 21 1/3 square units from 64 square units, we get 42 2/3 square units.

This tells us that you've found the area on the wrong side of the parabola! In other words, you found the area for the other part of the 4-unit by 16-unit rectangle.

Do you "see" how to fix your integrand?

 

[red and white kop!]

not really. i understand that i calculated the wrong area, but thats it. i dont get how to define the right one. can you explain?

 

[mmm4444bot]

… can you explain?

I just did.

Can you draw a picture? Integrating x^2 gives you the area UNDER the curve.

With each of the infinitely thin rectangles we're summing in the integration, you want their heights to be the distance from the parabola TO the line y = 16.

You're working with the heights being the distance from the parabola TO the x-axis, instead. That's the problem. All of your infinite rectangles are on the wrong side.

In other words, you want the rectangles ABOVE the curve because THAT's the region whose area you're summing.

Draw a picture of the given info.

Draw one representative rectangle.

It's width is dx.

What is the expression for its height?

In other words, what is the distance from the line y = 16 to any point on the parabola?

Did I say, "Draw a picture"?