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Back to FlashcardsMathematics / AP Calculus AB: 10.3.2 Finding Areas by Integrating with Respect to y: Part Two

AP Calculus AB: 10.3.2 Finding Areas by Integrating with Respect to y: Part Two

Mathematics8 CardsCreated 8 months ago

This section explains how to compute the area of a region using horizontal rectangles and integrating with respect to y, especially when the curves are more naturally expressed as functions of y (i.e., x in terms of y).

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Finding Areas by Integrating with Respect to y: Part Two

Area between two curves defined in terms of y:
• When finding the area of a region:
1. Sketch the region.
2. Determine how the rectangles will stack.
3. Find where the curves intersect.
4. Set up the integral.
5. Evaluate.

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Key Terms

Term
Definition

Finding Areas by Integrating with Respect to y: Part Two

Area between two curves defined in terms of y:
• When finding the area of a region:
1. Sketch the region.
2. Determine how the rectangles ...

note

  • To integrate with respect to y, the expression for the
    dimensions of the arbitrary rectangles must also be in terms of y.

Find the area of the region bound by y = 1/x, y = 1, y = 2, and the y-axis.

ln 2

Find the area of the region bound by y 2 = x, y = 2, y = −1, and y = x + 1.

4 1/2

Find the area of the region bound by x = 2sin y, x = −sin y, y = π/4
y = 3π/4

6/√2

Find the area of the region bound by the curve x = −y 2 and x = −3.

A=4√3

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AP Calculus AB: 10.3.2 Finding Areas by Integrating with Respect to y: Part Two
TermDefinition

Finding Areas by Integrating with Respect to y: Part Two

Area between two curves defined in terms of y:
• When finding the area of a region:
1. Sketch the region.
2. Determine how the rectangles will stack.
3. Find where the curves intersect.
4. Set up the integral.
5. Evaluate.

note

  • To integrate with respect to y, the expression for the
    dimensions of the arbitrary rectangles must also be in terms of y.

  • Notice that the height of the rectangle is equal to a small change in y. The base is given by the difference in the two x-values, which are expressed in terms of y.

  • Notice that the formulas for area use the same basic idea, integrating the area of a rectangle. The dimensions depend on how you define the rectangle.

  • Here is an example.

  • Start by sketching the graphs.

  • If you try to stack vertical rectangles, you will have rectangles that are defined by the difference of the same curve.

  • Instead, use horizontal rectangles.

  • Once you decide how you are going to stack, you must find where the stacking begins and ends. Look for the smallest value used for the variable of integration. That’s your lower limit of integration. The greatest value is the upper limit of integration.

  • Now set up the integral. The height is a small change in y. The base is the difference of the two expressions.

  • Setting up the integral is the trickiest part of this problem. Actually solving the integral is very straightforward.

Find the area of the region bound by y = 1/x, y = 1, y = 2, and the y-axis.

ln 2

Find the area of the region bound by y 2 = x, y = 2, y = −1, and y = x + 1.

4 1/2

Find the area of the region bound by x = 2sin y, x = −sin y, y = π/4
y = 3π/4

6/√2

Find the area of the region bound by the curve x = −y 2 and x = −3.

A=4√3

Find the area of the region bound by y2 = x and y = x − 2.

4 1/2

Find the area of the region bound by the curves x = y 2 and x = y 3 in the first quadrant.

A=1/12