Question: Find the volume out off from the paraboloid $x^2+\frac{y^2}{4}+z=1$ by the plane z = 0.
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Subject : Applied Mathematics 2

Topic : Triple integration and Applications of Multiple integrals

Difficulty : High

am2(82) • 337 views
 modified 10 months ago by Juilee ♦ 2.1k written 12 months ago by
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The xy plane cut the parabolid in the ellipse $x^2 + \frac{y^2}{4} = 1$

Hence the total volume

$v = \iint_R z \hspace{0.1cm}dx\hspace{0.1cm}dy\\ \hspace{0.1cm}=\iint_R\Big(1 - x^2-\frac{y^2}{4}\Big)dx \hspace{0.1cm}dy$

Where R i the area of the ellipse

$V = \int^1_{-1}\int^{2\sqrt{1-x^2}}_{-2\sqrt{1-x^2}}\Big(1 - x^2 - \frac{y^2}{4}\Big)dx \hspace{0.1cm}dy\\ \hspace{0.1cm} = 4\int^1_0\int^{2\sqrt{1-x^2}}_0 \Big(1 - x^2 - \frac{y^2}{4}\Big)dx \hspace{0.1cm}dy\\ \hspace{0.1cm}=4\int_0^1\Big[(1-x^2)y-\frac{y^3}{12}\Big]_0^{2\sqrt{1-x^2}}dx\\ \hspace{0.1cm}= 4\int_0^1\frac{4}{3}(1-x^2)^{3/2}dx = \frac{16}{3}\int_0^1(1-x^2)^{3/2}dx\\ \text{Put} \hspace{0.1cm} x = sin \theta, dx=cos\theta \hspace{0.1cm}d \theta\\ \hspace{0.1cm}=\frac{16}{3}\int_0^{\pi/2}(1-sin^2 \theta)^{3/2} cos \theta \hspace{0.1cm}d \theta = \frac{16}{3} \int_0^{\pi/2}cos^4\theta \hspace{0.1cm} d \theta\\ \hspace{0.1cm}=\frac{16}{3}\times \frac{1}{2}\beta(\frac{5}{2},\frac{1}{2})=\frac{8}{3}\frac{\sqrt{5/2}\sqrt{1/2}}{\sqrt{3}}\\ \hspace{0.1cm} = \frac{8/3 \times 3/2\times 1/2 \sqrt{\pi}\times\sqrt{\pi}}{2}\\ \hspace{0.1cm} = \pi$