Page 79 - Calc for the Clueless

Sometimes it becomes advantageous to set up the rectangles horizontally. We have h(y) and j(y), where j(y) is
always larger than or equal to h(y). The picture would look like this:

The area of each rectangle would again be the base times the height = [j(y k) - h(y k)] ∆y k. If we again properly
take the limits, we get

Example 2—

Find the area between the curves x = y and x = 3y + 4.
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If we try to make the rectangles as before, the top curve is the parabola, which is OK, but if you look very
closely there are two parts of the bottom curve, a little part of the bottom of the parabola and the straight line.
We would have to split the region and set up two integrals to find the area. However, the right curve is always
the straight line and the left curve is always the parabola. We take right-curve-minus-left-curve dy. To find the
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lowest and highest y values, we again set the curves equal to each other, y = 3y + 4, y - 3y - 4 = 0, (y + 1)(y-
4)= 0. y = -1 and y = 4.

Sometimes splitting the region cannot be avoided. We will do a problem that must be split no matter which way
you draw the rectangles. We will do it two ways. With experience you may be able to tell which way is better
before you begin the problem.

Example 3—

Find the area of the region bounded by y = x to the point (2,2), then the curve y = 6 - x from (2,2) to (4,-10),
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and then y = -5x/2 from (4,-10) to (0,0) where it meets the curve y = x.