Asked by GM
Sketch the region enclosed by the given curves.?
Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=3(x^(1/2)) , y=4, and 2y+3x=6
I have been working on this problem for the past 3 hours with a friend and we have just hit a brick wall. We have literally done everything we possible can and cannot figure this out. Please help!!!
Thank you!
Decide whether to integrate with respect to x or y. Then find the area of the region.
2y=3(x^(1/2)) , y=4, and 2y+3x=6
I have been working on this problem for the past 3 hours with a friend and we have just hit a brick wall. We have literally done everything we possible can and cannot figure this out. Please help!!!
Thank you!
Answers
Answered by
drwls
Draw the three curves
y1 = 4, y2 = -3/2 x +3 and y3 = (3/2)x^1/2
on the same graph to see what kind of a region you are dealing with.
The top border of the region is the horizontal y = 4 line extending from x = -2/3 to x = 64/9
There are two other bordering lines.
One is a straight line with negative slope extending downward from (-2/3, 4) to (1, 3/2). There is intersects the third line, which is the upper branch of a horizontal parabola extending from (1, 3/2) to (64/9, 4)
You can integrate either along x or y, but you would be integrating different functions. It looks a bit easier if you integrate it as the sum of two areas:
4 - (-3/2 x + 3) = 3/2 x +1 from x = -2/3 to 1, and
4 - (3/2)x^1/2 from x = 1 to x = 64/9
y1 = 4, y2 = -3/2 x +3 and y3 = (3/2)x^1/2
on the same graph to see what kind of a region you are dealing with.
The top border of the region is the horizontal y = 4 line extending from x = -2/3 to x = 64/9
There are two other bordering lines.
One is a straight line with negative slope extending downward from (-2/3, 4) to (1, 3/2). There is intersects the third line, which is the upper branch of a horizontal parabola extending from (1, 3/2) to (64/9, 4)
You can integrate either along x or y, but you would be integrating different functions. It looks a bit easier if you integrate it as the sum of two areas:
4 - (-3/2 x + 3) = 3/2 x +1 from x = -2/3 to 1, and
4 - (3/2)x^1/2 from x = 1 to x = 64/9
Answered by
drwls
Always start a problem of this kind by sketching the region. This is something you will have to do yourself.
Answered by
GM
Thank you so much drwls!!! You do not understand how much I appreciate your help! You are a lifesaver! :) I not only got the answer, but also understand how to do it now.