Volume

Calculus Level 5

Calculate the volume of the solid defined by the inequalities below:

z 3 x 2 + 2 y 2 , 3 x 2 + 2 y 2 + 5 z 2 1 z \geq 3x^{2} + 2y^{2}, \ \ 3x^{2} + 2y^{2} + 5z^{2} \leq 1


The answer is 0.1035.

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1 solution

Volume [ ImplicitRegion [ z 3 x 2 + 2 y 2 3 x 2 + 2 y 2 + 5 z 2 1 0.35 x 0.35 0.43 y 0.43 0. z 0.45 , { x , y , z } ] , Method Integrate ] 0.103495919745 \text{Volume}\left[\text{ImplicitRegion}\left[z\geq 3 x^2+2 y^2\land 3 x^2+2 y^2+5 z^2\leq 1\land -0.35\leq x\leq 0.35\land -0.43\leq y\leq 0.43\land 0.\leq z\leq 0.45,\{x,y,z\}\right], \\ \text{Method}\to \text{Integrate}\right] \Longrightarrow 0.103495919745 .

This is a ellipsoid cap over an elliptic paraboloid cup. By scaling the x x and y y coordinates to make the horizontal cross section circular, the integrations become much simpler. The volume can be unscaled back to the original dimensions afterwards linearly.

The cap: 1 10 ( 21 1 ) 1 5 ( 1 5 z 2 ) d z 5 ( 1 15 5 ( 21 1 ) 3 3000 ) + 1 10 ( 1 21 ) + 1 5 \int_{\frac{1}{10} \left(\sqrt{21}-1\right)}^{\frac{1}{\sqrt{5}}} \left(1-5 z^2\right) \, dz \Longrightarrow -5 \left(\frac{1}{15 \sqrt{5}}-\frac{\left(\sqrt{21}-1\right)^3}{3000}\right)+\frac{1}{10} \left(1-\sqrt{21}\right)+\frac{1}{\sqrt{5}} .

The cup: 0 1 10 ( 21 1 ) q d q 1 200 ( 21 1 ) 2 \int_0^{ \frac{1}{10} \left(\sqrt{21}-1\right)} q \, dq \Longrightarrow \frac{1}{200} \left(\sqrt{21}-1\right)^2 .

Adding the cap and cup, multiplying by for the "circular" cross section and unscaling; π ( 1 200 ( 21 1 ) 2 5 ( 1 15 5 ( 21 1 ) 3 3000 ) + 1 10 ( 1 21 ) + 1 5 ) 2 3 ( 40 5 21 21 + 31 ) π 300 6 0.103495912204 \frac{\pi \left(\frac{1}{200} \left(\sqrt{21}-1\right)^2-5 \left(\frac{1}{15 \sqrt{5}}-\frac{\left(\sqrt{21}-1\right)^3}{3000}\right)+\frac{1}{10} \left(1-\sqrt{21}\right)+\frac{1}{\sqrt{5}}\right)}{\sqrt{2} \sqrt{3}} \Longrightarrow \\ \frac{\left(40 \sqrt{5}-21 \sqrt{21}+31\right) \pi }{300 \sqrt{6}} \Longrightarrow \\ 0.103495912204

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