$$
\int_{-1}^{1} \int_{1-\sqrt{1-y^{2}}}^{1+\sqrt{1-y^{2}}} d x d y
$$

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) $\int_{0}^{2} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} d y d x$

Question

$$
\int_{-1}^{1} \int_{1-\sqrt{1-y^{2}}}^{1+\sqrt{1-y^{2}}} d x d y
$$

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) $\int_{0}^{2} \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} d y d x$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To switch the bounds of the double integral from $dx dy$ to $dy dx$, we need to visualize the region of integration in the xy-plane. ‖ ‖ 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.⚹◊What is the formula for changing the order of integration for a double integral in mathematics?⍭ Generate me a similar question◊

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