Question

Sketch the region bounded by the curves y = x^2 and y = 2 then set up the integral needed to find the volume of the solid generated by revolving this region about the x-axis using the shell method. a) integral_0^2 2 pi y^3/2 dy b) integral_0^2 4 pi/y^3/2 dy c) integral_0^2 2 pi y^2 dy d) integral_-Squareroot 2^Squareroot 2 pi y^3/2 dy e) integral_-Squareroot 2^Squareroot 2 4 pi y^3/2 dy

Answer 1

Answer:

Option b

Step-by-step explanation:

given that a region is bounded by the curves $y = x^2 \\y = 2$

It is rotated about x axis

We are to use shell method

Region picture is enclosed

We are to use shell method

Due to symmetry we can find form 0 to 2 and double it.

Here limits for y are from  0 to 2

The variable x can be written as

$x=\sqrt{y}$

A(y) = area = $2\pi (radius)(width)\\= 2\pi (y) (\sqrt{y} )$

Volume = $2\int\limits^2_0 {2\pi y\sqrt{y} } \, dy\\=4\pi\int\limits^2_0 y^{\frac{3}{2} } dy$

Hence option b is right

volume of the solid