Use a double integral to find the solid bounded by z = 1 - y^2 and z = y^2 -1 for 0 < x < 2
kvnmurty:
where is the given series? question lacks data
In place of previous question I posted another one
answer 16/3
Dont know
i am telling that the answer is 16/3. the limits of integration are -1 to 1. please see, I modified.
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In the y-z plane, the curve z = 1 - y² and the curve z = y² - 1 form a bounded area around y axis. It is symmetric wrt y axis. The shape of curve is parabolic.
In the y-z plane the intersection of z = 1 - y² and z = y² - 1 gives the limits of integration and define the bounded area.
z = 1 - y² = y² - 1 => 2 y² = 2 => y = +1 or -1
So the cross section area is bounded between y = -1 and y = +1.
In the x-axis direction the solid has a uniform area. So the solid is a cylinder with the length in the x direction with cross section in the y-z plane.
Volume = cross sectional area * length
![Volume = \int\limits^2_0 {Cross-section-Area} \, dx = \int\limits^2_0 ({ \int\limits^a_b {(z_2-z_1)} \, dy }) \, dx \\ \\\int\limits^2_0 ({ \int\limits^1_{-1} {(1-y^2-(y^2-1))} \, dy }) \, dx \\ \\=2\int\limits^2_0 { \int\limits^1_{-1} {(1-y^2)} \, dy } \, dx \\ \\2 \int\limits^2_0 {[y-y^3/3]_{-1}^1} \, dx =2 \int\limits^2_0 {[1-1/3-(-1+1/3)]} \, dx\\ \\=2 *\frac{4}{3} \int\limits^2_0 {1} \, dx\\ \\=2*\frac{4}{3}*[x]_0^2=2*\frac{4}{3}*2 = 16/3\\](https://tex.z-dn.net/?f=Volume+%3D+%5Cint%5Climits%5E2_0+%7BCross-section-Area%7D+%5C%2C+dx+%3D+%5Cint%5Climits%5E2_0+%28%7B+%5Cint%5Climits%5Ea_b+%7B%28z_2-z_1%29%7D+%5C%2C+dy+%7D%29+%5C%2C+dx+%5C%5C+%5C%5C%5Cint%5Climits%5E2_0+%28%7B+%5Cint%5Climits%5E1_%7B-1%7D+%7B%281-y%5E2-%28y%5E2-1%29%29%7D+%5C%2C+dy+%7D%29+%5C%2C+dx+%5C%5C+%5C%5C%3D2%5Cint%5Climits%5E2_0+%7B+%5Cint%5Climits%5E1_%7B-1%7D+%7B%281-y%5E2%29%7D+%5C%2C+dy+%7D+%5C%2C+dx+%5C%5C+%5C%5C2+%5Cint%5Climits%5E2_0+%7B%5By-y%5E3%2F3%5D_%7B-1%7D%5E1%7D+%5C%2C+dx+%3D2+%5Cint%5Climits%5E2_0+%7B%5B1-1%2F3-%28-1%2B1%2F3%29%5D%7D+%5C%2C+dx%5C%5C+%5C%5C%3D2+%2A%5Cfrac%7B4%7D%7B3%7D+%5Cint%5Climits%5E2_0+%7B1%7D+%5C%2C+dx%5C%5C+%5C%5C%3D2%2A%5Cfrac%7B4%7D%7B3%7D%2A%5Bx%5D_0%5E2%3D2%2A%5Cfrac%7B4%7D%7B3%7D%2A2+%3D+16%2F3%5C%5C "Volume = \int\limits^2_0 {Cross-section-Area} \, dx = \int\limits^2_0 ({ \int\limits^a_b {(z_2-z_1)} \, dy }) \, dx \\ \\\int\limits^2_0 ({ \int\limits^1_{-1} {(1-y^2-(y^2-1))} \, dy }) \, dx \\ \\=2\int\limits^2_0 { \int\limits^1_{-1} {(1-y^2)} \, dy } \, dx \\ \\2 \int\limits^2_0 {[y-y^3/3]_{-1}^1} \, dx =2 \int\limits^2_0 {[1-1/3-(-1+1/3)]} \, dx\\ \\=2 *\frac{4}{3} \int\limits^2_0 {1} \, dx\\ \\=2*\frac{4}{3}*[x]_0^2=2*\frac{4}{3}*2 = 16/3\\")
In the y-z plane the intersection of z = 1 - y² and z = y² - 1 gives the limits of integration and define the bounded area.
z = 1 - y² = y² - 1 => 2 y² = 2 => y = +1 or -1
So the cross section area is bounded between y = -1 and y = +1.
In the x-axis direction the solid has a uniform area. So the solid is a cylinder with the length in the x direction with cross section in the y-z plane.
Volume = cross sectional area * length
thanx n u r welcom
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