Question

Math

Posted 4 months ago

```
$\int_{-1}^{1} \int_{1-\sqrt{1-y^{2}}}^{1+\sqrt{1-y^{2}}} d x d y$
Switch the bounds of the double integral.
Choose 1 answer:
(A) $\int_{0}^{2} \int_{-\sqrt{1-(x-1)^{2}}}^{\sqrt{1-(x-1)^{2}}} d y d x$
(B) $\int_{0}^{2} \int_{-\sqrt{4-(x-1)^{2}}}^{\sqrt{4-(x-1)^{2}}} d y d x$
(C) $\int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} d y d x$
(D) $\int_{0}^{2} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} d y d x$
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 2

The given bounds for x are from $1-\sqrt{1-y^2}$ to $1+\sqrt{1-y^2}$. This describes a circle with radius 1 centered at (1,0)

step 3

To find the new bounds for y, we consider the projection of the circle onto the x-axis, which is the interval [0,2]

step 4

For a given x in [0,2], the corresponding y-values range from the bottom half to the top half of the circle. The equation of the circle is $(x-1)^2+y^2=1$

step 5

Solving for y, we get $y=\pm\sqrt{1-(x-1)^2}$. These are the new bounds for y

step 6

The correct double integral with switched bounds is $\int_{0}^{2} \int_{-\sqrt{1-(x-1)^2}}^{\sqrt{1-(x-1)^2}} dy dx$

A

Key Concept

Switching the order of integration in double integrals

Explanation

When switching the order of integration, it's important to understand the geometry of the region of integration and to express the limits for the new order correctly.

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