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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!

  • Calculus -

    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

  • Calculus -

    Always start a problem of this kind by sketching the region. This is something you will have to do yourself.

  • Calculus -

    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.

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