Let R be the region in the 1st quadrant of the xy-plane bounded by the graphs of

Question

Let  R be the region in the 1st quadrant of the xy-plane bounded by the graphs of y=2x^2 , y=3-x and y=0

Set up an integral representing the volume of the solid obtained by revolving R about each of the following lines. Do not evaluate the integral. Also, indicate which method you are using (washer/disk or shell).

a) The x-axis

b) The y-axis

c) The line x= -1

d) The line y= -1

e) The line x= 3

f) About the line y= 3

Let R be the region in the 1st quadrant of the xy-plane bounded by the graphs of y=2x^2 , y=3-x and y=0 Set up an integral representing the volume of the solid obtained by revolving R about each of the following lines. Do not evaluate the integral. Also, indicate which method you are using (washer/disk or shell). The x-axis The y-axis The line x= -1 The line y= -1 The line x= 3 About the line y= 3

Explanation / Answer

For integrals about y = constant, we use the disk/washer method, and for integrals about x = constant, we use the shell method because all functions given are y as a function of x.

Note that y = 2x^2 and y = 3 - x meet when

2x^2 = 3 - x or

2x^2 - (3 - x) = 0 or

2x^2 - 3 + x = 0 or

2x^2 + x - 3 = 0 or

(2x+3)(x-1) = 0

This equals 0 when x = -3/2 or x = 1

The x = -3/2 will not be relevant here, but x = 1 will be. Our region thus has y = 2x^2 from 0 to 1, y = 3 - x from 1 to 3, and y = 0 from x=0 to 3.

a) Rotating about the x-axis, or y = 0, we have the washer method.

Using I[a,b] for the integral from a to b, we have

I[0,3]pi f^2(x) dx, which becomes

I[0,1] pi (2x^2)^2 dx + I[1,3] pi(3-x)^2 dx or

I[0,1] 4 pi x^4 dx + I[1,3] pi (3-x)^2 dx

b) The y-axis

We now use the shell-method I[0,3] 2 pi x f(x) dx =

I[0,1] 2 pi x (2x^2) dx + I[1, 3] 2 pi x (3-x) dx or

I[0,1] 4 pi x^3 dx + I[1, 3] 6 pi x - 2 pi x^2 dx

c) The line x = -1

We again use the shell method. However, now the radius becomes x+1 rather than x.

I[0,3] 2 pi (x+1) f(x) dx =

I[0,1] 2 pi (x+1) 2x^2 dx + I[1, 3] 2 pi (x+1) (3-x) dx or

I[0,1] 4 pi x^3 + 4 pi x^2 dx + I[1, 3] -2 pi x^2 + 4 pi x + 6 pi dx

d) y = -1

We now use the disk/washer method (don't forget that we need to include the y = 0 side)

Now, the radius becomes f(x) + 1

Thus, we have I[0,1] pi (2x^2+1)^2 dx + I[1, 3] pi (4-x)^2 dx - I[0, 3] pi 1^2 dx

You may also express this as

I[0,1] pi (2x^2+1)^2 - pi 1^2 dx + I[1, 3] pi (4-x)^2 - pi 1^2 dx

e) About x = 3. We again use the shell method. Note that the radius is now 3 - x, not x - 3

Thus, we have I[0,3] 2 pi (3-x) f(x) dx =

I[0,1] 2 pi (3-x) 2x^2 dx + I[1, 3] 2 pi (3-x)(3-x) dx or

I[0,1] 12 pi x^2 - 4pi x^3 dx + I[1, 3] 2 pi (3-x)^2 dx

f) y = 3

Again, we use the disk/washer method. Now, the radius becomes 3 - f(x). Now, the y-axis actually creates the greater radius.

Thus, we have

I[0, 3] 3^2 pi dx - I[0,1] pi (3 - 2x^2)^2 dx - I[1, 3] pi (3 - (3-x))^2 dx or

I[0, 3] 9 pi dx - I[0,1] pi (3 - 2x^2)^2 dx - I[1, 3] pi x^2 dx

We may also write this as

I[0, 1] 9 pi - pi(3-2x^2)^2 dx + I[1, 3] 9 pi - pi x^2 dx

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