Question

Find the volume of the region bounded above by the paraboloid z=x^ 2 + y ^ 2 and below by the square : - 1 <= x <= 1, - 1 <= y <= 1 .​

Answer 1

Answer:

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Answer 2

Tip:

• If $g(x)\geq 0$ on the interval $[a,b]$ , then (double integration) $\iint\limits_R {f(x,y) \, dA$ has similar interpretation in terms of volume.

Explanation:

• Given $z=x^2+y^2$ and $R:~-1\leq x\leq 1,~-1\leq y\leq 1$.
• We have to find the  volume of the region bounded .
• We will solve this question by using the formula.

Step

Step 1 of 2:

Volume $=\iint\limits_R {z} \, dA$

where,  $z=x^2+y^2$ and $R:~-1\leq x\leq 1,~-1\leq y\leq 1$.

Step 2 of 2:

Volume $=\int_{-1}^1\int_{-1}^1 x^2+y^2 \, dxdy$

             $=\int_{-1}^1{(\frac{x^3}{3}+y^2x)|_{-1}^1} \, dy\\\\=\int_{-1}^1{(\frac{1}{3}+y^2-[-\frac{1}{3}-y^2])} \,dy\\\\=\int_{-1}^1{(\frac{2}{3}+2y^2)}\, dy\\\\=\frac{2y}{3}+\frac{2}{3}y^3\left|_{-1}^1$

             $=\frac{2}{3}+\frac{2}{3}-[-\frac{2}{3}-\frac{2}{3}]\\\\=\frac{4}{3}+\frac{4}{3}\\\\=\frac{8}{3}$

Final Answer:

The volume of the region is $\frac{8}{3}$