The equation of ellipsoid is $$x^2+\bigg(\frac{y}{2}\bigg)^2+\bigg(\frac{z}{c}\bigg)^2=1$$ I have taking the limits of integration $$\int_{0}^{1}\int_{-2}^{2}\int_{0}^{c\sqrt{1-x^2-\frac{y^2}{4}}}dzdydx$$ Am I right just confused little bit. That's why I need confirmation. The second question is how can I change the order of integration to evaluate the following
- $\int_{0}^{4}\int_{0}^{1}\int_{2y}^{2} \frac{4\cos(x^2)}{2\sqrt{z}}dxdydz$
- $\int_{0}^{2}\int_{0}^{4-x^2}\int_{0}^{x} \frac{\sin2z}{4-z}dydzdx$
- $$\int_{0}^{1}\int_{3\sqrt{z}}^{1}\int_{0}^{\ln3} \frac{\pi e^{2x} \sin(\pi y^2)}{y^2}dxdydz$$
For (1)
With y-innermost:
Since $0\leq y\leq 1,$ $2y\leq x\leq 2$ and $0\leq z\leq 4$
The bounds of $y$-are $0\leq y\leq \frac{x}{2}$, the z limits are unaffected by $y$ and $x$,
hence
$$\int_{0}^{4}\int_{0}^{2}\int_{0}^{\frac{x}{2}}dy dx dz.$$
Solving this I get
$$\int_{0}^{4}\int_{0}^{2}\int_{0}^{\frac{x}{2}}dydxdz=\frac{8\sin4}{3}$$
Am I right changing in order of integration?
For (2) With y-innermost: since the limits of integration are $0\leq x\leq2,$ $0\leq z\leq 4-x^2$ and $0\leq y\leq x$, then I get the
$$\int_{0}^{2}\int_{0}^{2}\int_{0}^{\sqrt{4-z}}\frac{\sin2z}{4-z}dy dx dz$$
When I try to solve this (2) after changing the order of integration I am not still getting the integral in form of elementary functions, Can anyone help!
For (3) With z-innermost: I get
$$\int_{0}^{\ln3}\int_{0}^{1}\int_{0}^{\frac{y^2}{9}}\frac{\pi e^{2x} \sin(\pi y^2)}{y^2}dzdydx$$but still after changing the order of integration in (3) I can't simplified this.
Please help in solving all these four!
Thanks in advance!