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Sử dụng tích phân bội ba tính thể khối nằm bên trong mặt trụ x^2 +y^2 = 4, bên dưới mặt z= 10−x^2−y^2 và bên trên mặt z=−căn(2$) ta được kết quả aπ. Tìm a Chú ý kết quả điền vào là số thập phân làm tròn đến 3 chữ số sau dấu phẩy.
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Answer

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Step 1:
This problem is asking us to calculate the volume of the solid that lies inside the cylinder x^2 + y^2 = 4, below the surface z = 10 - x^2 - y^2 and above the surface z = -\sqrt{4x^2 + 4y^2} using triple integration.

The result will be in the form of $$a\pi$$, and we need to find the value of $$a$$ rounded to three decimal places.

Step 2:
: Set up the integral

The limits for $$z$$ are given by the two surfaces, which we need to convert to cylindrical coordinates: $$-\sqrt{4r^2} \leq z \leq 10 - r^2$$.
To set up the triple integral, we need to determine the limits of integration. Since we are dealing with a cylinder, it is easier to use cylindrical coordinates.

Step 3:
: Write down the triple integral

\int_{0}^{2\pi} \int_{0}^{2} \int_{-\sqrt{4r^2}}^{10 - r^2} r \, dz \, dr \, d\theta
The factor of r comes from the Jacobian determinant of the transformation from Cartesian to cylindrical coordinates. The limits of integration are those determined in Step 1. So the triple integral becomes:

Step 4:
: Evaluate the integral

This gives us $$\frac{2}{3} \cdot [\theta]_{0}^{2\pi} = \frac{4\pi}{3}$$.
This gives us r \cdot [z]_{-\sqrt{4r^2}}^{10 - r^2} = r \cdot [(10 - r^2) - (-\sqrt{4r^2})] = r \cdot (10 - r^2 + 2r).

Final Answer

The volume of the solid is \frac{4\pi}{3}, so a = 1.333 when rounded to three decimal places.