微积分2|Calculus 2 MAST10006代写

This subject will extend knowledge of calculus from school. Students are introduced to hyperbolic functions and their inverses, the complex exponential and functions of two variables. Techniques of differentiation and integration will be extended to these cases. Students will be exposed to a wider class of differential equation models, both first and second order, to describe systems such as population models, electrical circuits and mechanical oscillators. The subject also introduces sequences and series including the concepts of convergence and divergence.

这是一份unimelb墨尔本大学MAST10006的成功案例

微积分2|Calculus 2 MAST10006代写 unsw


问题 1.

Find the mass of a ball 9 of radius $a$ whose density is numerically equal to the distance from a fixed diametral plane.

证明 .

Let the ball be the inside of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$, and let the fixed diametral plane be $z=0$. Then $M=\iiint|z| d V$. Use the upper hemisphere and double the result. In spherical coordinates.
$$
\begin{aligned}
M &=2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{a} z \cdot \rho^{2} \sin \phi d \rho d \phi d \theta=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \int_{0}^{a} \rho \cos \phi \cdot \rho^{2} \sin \phi d \rho d \phi d \theta \
&\left.=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \frac{1}{4} \rho^{4} \cos \phi \sin \phi\right]{0}^{a} d \phi d \theta=\frac{1}{2} a^{4} \int{0}^{2 \pi} \frac{1}{2} \sin ^{2} \phi \int_{0}^{\pi / 2} d \theta=\frac{1}{4} a^{4} \int_{0}^{2 \pi} d \theta=\frac{1}{4} a^{4} \cdot 2 \pi=\frac{1}{2} \pi a^{4}
\end{aligned}
$$






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MAST10006 COURSE NOTES :

Find the surface area $S$ of the part of the sphere $x^{2}+y^{2}+z^{2}=4 z$ inside the paraboloid $z=x^{2}+y^{2}$.
$\square$ The region $\mathscr{R}$ under the spherical cap (Fig. 44-33) is obtained by finding the intersection of $x^{2}+y^{2}+z^{2}=4 z$ and $z=x^{2}+y^{2}$. This gives $z(z-3)=0$. Hence, the paraboloid cuts the sphere when $z=3$, and $\mathscr{B}$ is the disk $x^{2}+y^{2} \leq 3 . \quad S=\iint_{a} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} d A . \quad 2 x+2 z \frac{\partial z}{\partial x}=4 \frac{\partial z}{\partial x}, \quad \frac{\partial z}{\partial x}=-\frac{x}{z-2} . \quad$ Similarly, $\frac{\partial z}{\partial y}=-\frac{y}{z-2}$. Hence,
$$
1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=1+\frac{x^{2}}{(z-2)^{2}}+\frac{y^{2}}{(z-2)^{2}}=\frac{(z-2)^{2}+x^{2}+y^{2}}{(z-2)^{2}}=\frac{\left(x^{2}+y^{2}+z^{2}\right)-4 z+4}{(z-2)^{2}}=\frac{4}{(z-2)^{2}}
$$
Therefore,
$$
\left.S=\iint_{\pi} \frac{2}{z-2} d A=\int_{0}^{2 w} \int_{0}^{\sqrt{3}} \frac{2}{\sqrt{4-r^{2}}} r d r d \theta=-\int_{0}^{2 \theta} 2 \sqrt{4-r^{2}}\right]{0}^{\sqrt{3}} d \theta=-2 \int{0}^{2 \pi}(1-2) d \theta=2 \cdot 2 \pi=4 \pi
$$




















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