Asosiy tarkibga o'tish

6.5-bo'lim : Ko'proq ovoz balandligi muammolari

  1. \(r\) radiusli sharning hajmini aniqlash uchun ushbu qismdan hajmni topish usulidan foydalaning.
  2. Bazasi \(x = 2 - {y^2}\) va \(x = {y^2} - 2\) bilan chegaralangan mintaqa bo‘lgan va ko‘ndalang kesimlari asosi \(y\) o‘qiga perpendikulyar bo‘lgan kvadratlar bo‘lgan qattiq jismning hajmini toping. Ko'ndalang kesimlarning eskizini ko'rish uchun quyidagi rasmga qarang.
    There is a standard xy-axis system shown in this sketch. Also shown are the graphs of $x=2-y^{2}$ and $x=y^{2}-2$. Both graphs start at their respective vertexes and then go left/right respectively until they hit the y-axis and then stop. There is also a square that rises up out of the graph and is perpendicular to the positive y-axis. The bottom right/left points of the square are on the curves.
  3. Bazasi \(r\) radiusli disk va ko‘ndalang kesimlari balandligi asos uzunligining yarmiga teng, asosi \(x\) o‘qiga perpendikulyar bo‘lgan to‘rtburchaklar bo‘lgan qattiq jismning hajmini toping. Ko'ndalang kesimlarning eskizini ko'rish uchun quyidagi rasmga qarang (eskizda ijobiy \(x\) o'qi va musbat \(y\) o'qi ko'rsatilgan).
    There is a non-standard xy-axis system shown in this sketch. The positive y-axis is horizontal and goes to the right out of the origin while the positive x-axis is vertical and goes down out of the origin. A circle of radius r is also shown on the graph and it is noted that its equation is $x^{2}+y^{2}=r^{2}$. There is also a rectangle that rises up out of the graph and is perpendicular to the positive x-axis. The bottom right/left points of the rectangle are on the circle.
  4. Bazisi \(y = {x^2} - 1\) va \(y = 3\) bilan chegaralangan mintaqa va ko‘ndalang kesimlari asosi \(y\) o‘qiga perpendikulyar bo‘lgan teng yonli uchburchaklar bo‘lgan qattiq jismning hajmini toping. Ko'ndalang kesimlarning eskizini ko'rish uchun quyidagi rasmga qarang.
    There is a standard xy-axis system shown in this sketch. Also shown are the graphs of $t=x^{2}-1$ and $y=3$. The graph of the parabola starts at its vertex and goes up until it hits the graph of $y=3$. There is also a triangle that rises up out of the graph and is perpendicular to the positive y-axis. The base right/left points of the triangle are on the graph of the parabola.
  5. Bazasi \(x = 2 - {y^2}\) va \(x = {y^2} - 2\) bilan chegaralangan mintaqa va ko‘ndalang kesimlari \(y\) o‘qi markazida joylashgan aylananing yuqori yarmi bo‘lgan qattiq jismning hajmini toping. Ko'ndalang kesimlarning eskizini ko'rish uchun quyidagi rasmga qarang.
    There is a standard xy-axis system shown in this sketch. Also shown are the graphs of $x=2-y^{2}$ and $x=y^{2}-2$. Both graphs start at their respective vertexes and then go left/right respectively until they hit the y-axis and then stop. There is also a graph of the upper half of a circle that rises up out of the graph and is perpendicular to the positive y-axis. The right/left points of the diameter of the semi-circle are on the curves.
  6. Bazasi \(y = \cos \left( x \right)\) va \( - \frac{\pi }{2} \le x \le \frac{\pi }{2}\) orasidagi \(x\) o‘qi bilan chegaralangan hudud bo‘lgan “silindr”dan kesilgan takoz hajmini toping. Takozning yuqori va pastki qismlari orasidagi burchak \(\frac{\pi }{4}\). "Tsilindr" va xanjarning eskizi uchun quyidagi rasmga qarang (eskizda ijobiy \(x\) o'qi va musbat \(y\) o'qi ko'rsatilgan).
    This image has two parts. Both parts have a non-standard xy-axis system shown. The positive y-axis is horizontal and goes to the right out of the origin while the positive x-axis is vertical and goes down out of the origin. On the left is a 3D solid that looks almost like half of a cylinder. At the bottom of the solid is the part of the graph of $y=\cos(x)$ in the domain $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$. The “hump” of the graph is the rightmost point on the y-axis. The solid then rises out of this region and gives a shape that is like a half cylinder except its cross sections are not semi-circles but instead the region at the base. Also shown in the left sketch is the top of the wedge that is inside the solid and it is shown that the top of the wedge makes an angle of $\frac{\pi}{4}$ with the base of the solid. On the right side is just the wedge itself without the rest of the solid above it. The “pointed” edge of the wedge is on the x-axis and the rounded side of the wedge again is in the shape of the graph of $\cos(x)$. The top of the wedge makes an angle of $\frac{\pi}{4}$ with the base of the wedge.
  7. \(r\) radiusli sfera uchun \(\varphi \) burchagi bilan aniqlangan qopqoq hajmini toping, bu erda \(\varphi \) \(y\) o'qi tomonidan hosil qilingan burchak va qopqoqning bosh nuqtasidan pastki qismigacha bo'lgan chiziqdir. \(\varphi \) burchagi tasviri uchun quyidagi rasmga qarang.
    This is an image of the full sphere. It is “centered” at the origin of a typical xy-axis system. The y-axis rises out of the top of the sphere and the x-axis moves out of the right side of the sphere. The “cap” of the sphere is shown in a different color to illustrate it. There is a line drawn out of the center of the sphere, which is also the origin of the xy-axis system, to where the base of the cap intersects the sphere. The angle between the positive y-axis and this line is give as $\varphi $.