Textbook Question
Determine the area of the shaded region bounded by the curve x^2=y^4(1−y^3) (see figure).
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9. Graphical Applications of Integrals
Area Between Curves
2:47 minutesProblem 6.2.7
Textbook Question
Express the area of the shaded region in Exercise 5 as the sum of two integrals with respect to y. Do not evaluate the integrals.
Verified step by step guidance1
Identify the region bounded by the lines $y = 2 - x$ and $y = x$, and the vertical line $x = 1$. The vertices of the shaded triangular region are at points $(0, 2)$, $(1, 1)$, and $(1, 0)$.
Since the problem asks for the area expressed as the sum of two integrals with respect to $y$, we need to rewrite the boundary curves in terms of $x$ as functions of $y$. From $y = 2 - x$, solve for $x$: $x = 2 - y$. From $y = x$, solve for $x$: $x = y$.
Determine the range of $y$ values for the two parts of the region. The lower part of the region extends from $y = 0$ to $y = 1$, and the upper part extends from $y = 1$ to $y = 2$.
For $0 \leq y \leq 1$, the left boundary is $x = y$ and the right boundary is $x = 1$. So, the area contribution here is the integral $\int_0^1 (1 - y) \, dy$.
For $1 \leq y \leq 2$, the left boundary is $x = 0$ and the right boundary is $x = 2 - y$. So, the area contribution here is the integral $\int_1^2 (2 - y) \, dy$. The total area is the sum of these two integrals.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Setting up integrals with respect to y
When expressing area as integrals with respect to y, the region is sliced horizontally. This requires rewriting the boundary curves as functions of y (i.e., x in terms of y) and integrating over the y-intervals that cover the region. This approach is useful when the region is bounded more naturally by horizontal slices.
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Finding inverse functions of boundary curves
To integrate with respect to y, the given functions y = 2 - x and y = x must be inverted to express x as functions of y. For example, from y = 2 - x, we get x = 2 - y, and from y = x, we get x = y. These inverse functions define the horizontal boundaries of the region for integration.
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Partitioning the region for multiple integrals
The shaded region is divided into two parts along y = 1 because the left and right boundaries change at this point. Each part corresponds to a different pair of boundary functions in terms of y, so the total area is expressed as the sum of two integrals over different y-intervals.
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