Find the volume bounded by the $xy$ plane, cylinder $x^2 + y^2 = 1$ and sphere $x^2 + y^2 +z^2 = 4$.   I am struggling with setting up the bounds of integration.  First, I will calculate the 'first-quadrant' piece of the volume.  $z$  will traverse from $0$ to $2$.  $x$ should start from the cylinder and go to the edge of the current circle of the sphere: $$\sqrt{1-y^2} \le x \le \sqrt{4-x^2-z^2}$$ However, the same applies to $y$: (I am only calculating half of the volume right now, where the smaller circle is the lower bound):  $$\sqrt{1-x^2} \le y \le \sqrt{4-y^2-z^2}$$  However, this cannot work as both $x$ and $y$ are dependent.  What is the error?
120512
Find the volume of the region inside both the sphere $x^2+y^2+z^2=4$ and the cylinder $x^2+y^2=1$   $$2\int_0^{2\pi}\int_0^{\frac{\pi} 6}\int_0^{2}1.r^2\sin \varphi dr d\varphi d\theta + \int_0^{2\pi}\int_{\frac{\pi} 6}^{\frac{5\pi} 6}\int_0^{\frac 1 {\sin \varphi}}1.r^2\sin \varphi dr d\varphi d\theta $$    This question is in the lecture notes but I didn't understand how the answer is obtained. Could someone help?
91561
The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.  The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.  We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:  $$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$  where $F$ is the equation of the sphere.  Plugging in the expressions and simplifying ($z \ge 0)$, we get:  $$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$  Converting to polar coordinates, we have:  $$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$  Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I  going wrong?
61233
I am trying to show that the sphere $S^2$ and $\mathbb{R}^2$ are not homeomorphic.I understand that you can't 'compress' a 3D shape into a 2D plane but I don't know how I would express this formally.  $S^2 = \{(x, y, z) ∈ \mathbb{R}^3: x^2 + y^2 + z^2 = 1\}$  As always, any help is appreciated!
55800
How can i compute the volume of a cone whose length of its side is $R$ and the vertex of the cone forms an angle $2θ$ .  The top cone is a cap of a sphere of radius $R$.    I tried to solve first in 2 dimension and then use solid of revolution:    Some help please to find the volume of this cone.
40391
I have a complex integral $$\int_{|z|=r}x \, dz$$ for the positive portion of the circle.  I know that this integral seems easy enough but I am having trouble with it, and I'm fairly certain my answer is incorrect.  I wanted to use a parameter such that $x=r\cos\theta$, so we then have $dz=dx+i\,dy=dx$ since there is no reason to have have the $i\,dy$ since the imaginary part is equal to $0$.  So now I have the double integral  $$\int^r_0\int^\pi_0 r \cos\theta \, dx$$  which equals $0$ since integration the $\cos\theta$ gives $\sin\theta$ and evaluated from $0$ to $\pi$ makes the integral evaluate to $0$.  Am i misinterpreting part of the question?  Way off base?  The book suggests to also solve the problem by realizing $x=\frac{1}{2}\left(z+\frac{r^2}{z}\right)$ along the circle.  For this portion of the problem I assume that I would integrate the above by $z$ as if it were a normal calculus problem, but I have problems with what to put as upper and lower limits.  Thanks in advance!
84490
Compute the area of that portion of the conical surface $x^2 + y^2 = z^2$ which lies between the two planes $z = 0$ and $x + 2z = 3$. Ans:$2\pi\sqrt{6}$. Thank you.
116788
Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$  Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side.  How we can solve such integrals?I need your help.  Without using the Gauss Divergence theorem.
61584
I want to evaluate the integral $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using the substitution $x=r\cos\theta$ and $y=r\sin\theta$ where the Jacobian is given as $|J(r, \theta)|=r$ and thus $dxdy=rdrd\theta$.  I need to set the range for $r$ and $\theta$. For $\theta$, it seems obvious that it ranges from $0$ to $\frac{\pi}{2}$ to cover all parts of the rectangle. But I am confused with the range for $r$. Would it be OK to set minimum $r$ as $0$ and maximum $r$ as $\sqrt{2}$, which is the diagonal of the square domain? I think I am confused with the fundamentals of integration.
24556
I am using two books for my calculus refresher.   Thomas' Calculus  Higher Math for Beginners by Ya. B. Zeldovich   My question is : When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.  Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??  Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.  In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.
71477
How do I find the volume when revolving the region bounded by $y=1-\frac{1}{2}x$, $y=0$, and $x=0$ about the line $ x=-2$?  Would it be $x=2-2y$ so radius $r(y) = 2-2y -(-2) $ => $r(y)= 4-2y$   $π\int (4-2y)^2 dy$ ?  What would be my limits of integration? Would it be from 0 to 2?? yes?
50099
Question: Let $R$ be the region in $xy$-plane bounded by the curves $y=x^2,y=4x^2,xy=1,xy=5$. Then find the value of line integral $\iint_R\frac{y^2}{x}dydx$.  I found it tough and time taking if I evaluate this integral from general method, is there any theorem which make it easy? Since the question is asked in competitive exam so I hope there must be an easy and fast way to solve it. Any help will be appreciable. Thanks in advance.
22559
Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions"   Is there a easy way to do this?   Maybe like FTOC II?
52820
The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume?   Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem?   Please help, thanks!
115026
Question: Sketch the complex region given by the inequality:  $$|z-1|\lt|z+1|$$  My attempt:  I tried to square both sides and use the fact that $z=x+iy$ however upon expanding I got a form that I couldn't recognise as any shape
47826
Evaluate the following integral $$\iint_S (x+y+z) \, dS$$  where $S$ is the surface of the cube $[0,1] \times [0,1] \times [0,1]$  Honestly, I don't know what to do. All I know is that you have to evaluate the integral for each of the 6 surfaces of the cube. The normals for them I can do and the equations of the planes, but other than that I couldn't do much.  Can someone break it down for me, please?  PS: No divergence theorem yet for me.
85888
Ted Shifrin on this site posed an interesting problem to me: show that   $$ \lim_{(x,y) \to (0,0)} \frac {|x|^ \alpha |y|^ \beta} {|x|^ \gamma + |y|^ \delta} \text {exists} \iff \frac\alpha\gamma + \frac\beta\delta &gt; 1, \,\,\,\,\text{where } \alpha, \beta, \gamma, \delta &gt;0$$  I think I've got the $(\Leftarrow)$ direction as follows: assume WLOG that $\gamma \leq \delta$. Then switch to polar coordinates and get   $$\frac {r^ {\alpha + \beta} (*) } {r^\gamma (1(*) + r^{\delta - \gamma}(*))},$$  where $(*)$ represents some trig stuff that is bounded near zero.   Edit: I need to make sure that $(1(*) + r^{\delta - \gamma}(*))$ is bounded below here.  The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
89341
The actual question was to find the area bounded by the curve $y=\int_{-1}^{1}(1-x^2)^ndx$ and coordinate axes. But I haven't came across these type of problems with power $n$.