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Evaluate the integralL" [ (- 1) min(1,V) dydr, where min(T,y) is the minimum value of and Let f,9 : R+ Rbe functions of one variable such that f" aId 9&qu...

Question

Evaluate the integralL" [ (- 1) min(1,V) dydr, where min(T,y) is the minimum value of and Let f,9 : R+ Rbe functions of one variable such that f" aId 9" are continuous_ Show that L Kco) 9" (y)) dyds f(0) 9(0) f(2) 9(2) + 2f' (2) "24 (0) . Let a > 0 In spherical coordinates. surface defined by r = 2acos for 0 < 0< 3 Find the volume of the solid enclosed by the surface, #S function of a Consider the inner product space P(R) with inner product(p. 4)p(r)alz) dr

Evaluate the integral L" [ (- 1) min(1,V) dydr, where min(T,y) is the minimum value of and Let f,9 : R+ Rbe functions of one variable such that f" aId 9" are continuous_ Show that L Kco) 9" (y)) dyds f(0) 9(0) f(2) 9(2) + 2f' (2) "24 (0) . Let a > 0 In spherical coordinates. surface defined by r = 2acos for 0 < 0< 3 Find the volume of the solid enclosed by the surface, #S function of a Consider the inner product space P(R) with inner product (p. 4) p(r)alz) dr Vp,q € P(R)_ Use the Gram-Schmidt process t0 construct orthonormal lutsis from the basis {LI,I} vOUI answer t0 part (a) . give the least sqpuares approximation in P(R) to the Using function f(c) on the interial [o. 1] (Hint: You may use the following result without proof: for n =1,2 ) 1& (-1)"(&xe n); where ao = Evaluate the integral L6 [ cosh(r") dr dy dz. and between che planes and Consider the solid inside the cylinder the centre of MLASX of the solid, 14 Let the density be given by p(I V:) Fiud



Answers

If $S$ is the surface defined by a function $z=f(x, y)$ that has continuous first partial derivatives throughout a region $R_{x y}$ in the $x y$ -plane (Figure $16.49 ),$ then $S$ is also the level surface $F(x, y, z)=0$ of the function $F(x, y, z)=f(x, y)-z$ . Taking the unit normal to $R_{x y}$ to be $\mathbf{p}=\mathbf{k}$ then gives
$$
\begin{aligned}|\nabla F|=\left|f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right| &=\sqrt{f_{x}^{2}+f_{y}^{2}+1} \\|\nabla F \cdot \mathbf{p}| &=\left|\left(f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right) \cdot \mathbf{k}\right|=|-1|=1 \end{aligned}
$$
and
$$
\iint_{R_{\mathrm{xy}}} \frac{|\nabla F|}{|\nabla F \cdot \mathbf{p}|} d A=\iint_{R_{\mathrm{xy}}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y
$$
Similarly, the area of a smooth surface $x=f(y, z)$ over a region $R_{y z}$ in the $y z$ -plane is
$$
A=\iint_{R_{y x}} \sqrt{f_{y}^{2}+f_{z}^{2}+1} d y d z
$$
and the area of a smooth $y=f(x, z)$ over a region $R_{x z}$ in the $x z$ -plane is
$$
A=\iint_{R_{\mathrm{xz}}} \sqrt{f_{x}^{2}+f_{z}^{2}+1} d x d z
$$
Use Equations $(11)-(13)$ to find the area of the surfaces in Exercises $39-44 .$
The surface in the first octant cut from the cylinder $y=(2 / 3) z^{3 / 2}$ by the planes $x=1$ and $y=16 / 3$

All right for these problems I have sketched the region of um you know integration were given to pull a cord to polar polar curves. You have to circle central origin the biggest one as articles one and the second polar curve is R equals one plus closing pita that is known as a cardio. It and it looks the test so R. Equals one for consideration based on that. It should be clear that the data values Would be ranging from -5 or two all the way to pirate tube. Um The range values for our horrible revenging from. So if you imagine starting at the origin and you know going through the regional integration you see that the region of integration to start at The polar curve. Archos one and it will end at the polar curve. Are it was one plus closing data. So The range values for our will go from 1-1 plus closing people again. The data values again start at negative poverty too because that's where the region starts. So if you start increasing the values for data, you know you will continue. The region will be included in that range of violence with data and it will end our data equals prior to right. So it started negative power to and have positive power to. And uh the range mothers receive well we know that The base of the solid would be on except by plane. So that would be z equals zero and it is pretty much Bounded on the top by the plane. Z. four. So that's pretty much force for in accordance as well. So that's it. We have all of the limits for integral. So are the setup for the iterated integral which is B. The integral from negative pi over two to buy or to The integral from 1 to 1 plus causing theater fix this. And then lastly the integral from 0 to 4 can be. We're not giving any um formula for the integral to function the integral. So we're just gonna leave this S. F. Of our committee to come up Z. And lastly behalf, busy times are as the are the data.

You have? Why two Vehicle to 4? My Nets said. So this implies for every respect to as a partial derivative. Oh, that's It's equal to zero. No pasha directly with respect to were zeke respect to set, It's equal to -1. So then this implies that the square root of F X squared plus F Z x squared plus one is going to be square roots of suit. So then this implies that our area our area will be called to the Dublin seeker over our X. Z Square root of two. The mm. And this is equal to Jessica From 0 to 2. Yeah, from 0-4. My next ex squared Square roots of two, the X D. Z. And this is equal to the square root of two as a constant integer from 0 to 2. You have four minutes, Z x squared the Z. And this is equal to 16 square which of two divided by three itself. I know

I should innovative of F. With respect to X. It's equal to X divided by square roots of X squared. That's why I squared the pressure derivative of F. With respect to why it's equal to why divided by square roots of X squared plus Y squared. So then this implies that the square root of f X squared plus F Y squared plus one. It's going to be equal to the square root of X squared. So this is going to be ex Quigg divided by square of this council cells. So you have X squared plus Y squared, you have bloods, Y squared divided by X squared plus y squared blacks one. So then this implies that our area area is going to be the Dublin secret over our This is going to this would be equal to the square root of two. So we have square roots of two. The X. The way. So then this is equal to Square root of two, It's a constant. So we bring that out. The integral over our the eggs the right. And this is the area between the ellipse in the cycle. So you didn't have Square roots of two. Then the area between between the ellipse in the cycle in the cycle. So this is going to be Equal to Square Roots of two. You want to be equal to the square root of two, 65 minus. Hi. And this is equal to five by Square root of two answer. Final answer. Mhm.

So why? Why? It's a call to Sue divided by three. Z. Two three divided by two. So then this implies that have with respect to eggs of eggs safe. It's equal to zero in every respect to zet eggs Z. It's equal to zero exponents 1 on two. So this implies that the square roots of F. X squared plus F. Z squared plus one will be equal to square roots of zero plus one. So then we have Why he called to 16 divided by three in place. You have 16 divided by three. To be equal to Sue divided by three zip Exponents three divided by two. So then this implies that are said it's equal to four soon area. Then is going to be the Dublin seek where. Mhm. So we have doubling. So you guys you have from 0 to 4. Then from 0 to 1 I have zero plus one. The X. Disease. To be equal to see how they to girl from 0 to 4. Zach plus one dizzied. And this is equal to this is equal to so divided by me five square roots for five minutes. One. As the final answer.


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