Let the plane graph be surrounded by curves y = X2, x = Y2 (1) The area of plane figure; (2) The volume of the body of revolution obtained by rotating the figure around the x-axis

Let the plane graph be surrounded by curves y = X2, x = Y2 (1) The area of plane figure; (2) The volume of the body of revolution obtained by rotating the figure around the x-axis

(1) Since the intersection point of the curve y = X2, x = Y2 is (0, 0), taking x as the integral variable, the
The area of the figure is:
（S＝∫
One
Zero
(
x−x2)dx＝(2
3x3
2−1
3x3)|
One
Zero
＝1
Three
(2) Volume of Revolution: VX = π∫
One
Zero
((
x)2−x4)dx
＝π∫
One
Zero
(x−x4)dy
＝π(1
2x2−1
5x5)|
One
Zero
＝3
10π

The area of the figure enclosed by the curves y = x 2 and x = y 2 is calculated, and the volume of the body of revolution is obtained by rotating it around the Y axis

Because the intersection points of the curves y = x2 and x = Y2 are 0 and 1,
Therefore, the enclosed area is integrated on (0, 1),
So there are:
A＝∫
One
Zero
(
x −x2)dx=[2
3x3
2−x3
3]
One
Zero
=1
Three
Because of one revolution around the y-axis, y is integrated, and the integration region is (0, 1),
Therefore, it can be concluded that:
V＝π∫
One
Zero
 (y−y4)dy=π[y2
2−y5
5]
One
Zero
=π3
10=3π
10．

Find the area of the plane figure surrounded by the curve y = x ^ 2 x = 1, y = 0, and the volume of the rotating body generated by the graph rotating around the X axis

Find the area of the plane figure surrounded by the curve y = x? 2, x = 1, y = 0, and the volume of the rotating body generated by the graph rotating around the X axis
Area s = [0,1] ∫ x? DX = x? / 3 [0,1] = 1 / 3
Volume v = [0,1] ∫ π y? DX = [0,1] ∫ π x ⁴ DX = π (x ^ 5) / 5 [0,1] = π / 5

Find the area of the plane figure surrounded by y = x ^ 2, y = x and the volume of the rotating body obtained by rotating around the X axis RTRTRT,3Q

(1) the area element Da (x) = (x-x ^ 2) DX of the figure at x ∈ [0,1], so the area obtained is a = ∫ [0,1] Da (x) = ∫ [0,1] (x-x ^ 2) DX = 1 / 6. (2) the volume element of the rotating body of the figure at x ∈ [0,1]

Find the area of the plane figure enclosed by the straight line y = x and the square of parabola y = X

It's obviously above the x-axis
y=√x
The intersection point is (0,0), (1,1)
The parabola is above
So s = ∫ (0 to 1) (√ x-x) DX
=2x √ X / 3-x ^ 2 / 2 (0 to 1)
=(2/3-1/2)-(0-0)
=1/6

How to find the area of a plane figure enclosed by a parabola y equal to the square of X and a straight line y equal to x plus 2

With definite integral, the integrand is x + 2-x ^ 2, and the integral interval is - 1 to 2. If I'm not wrong, the result is 9 / 2
Integral sign, lower limit-1, upper limit 2, integrand function x + 2-x ^ 2, and then DX, do integration, and then get 1 / 2x ^ 2 + 2x-1 / 3x ^ 3, and then add the upper limit 2 and the lower limit-1, and then subtract them. Will the following be calculated?
This place can only be written to this point, can not be more detailed
PS: I did it again. I didn't make a mistake. It was 9 / 2

Find the area of the figure enclosed by two parabola y = x 2 and y = 1

Because the intersection point of y = x2 and y = 1 is (± 1,1)
The area of the enclosed figure a = ∫
One
−1
(1−x2)dx=2∫
One
Zero
(1−x2)dx＝4
Three

The area of the figure (shadow part) enclosed by parabola y2 = x and straight line y = X-2 is () A. 9 Two B. 3 Two C. 7 Six D. 10 Three

y2＝x
Y = x − 2, x = 1, y = - 1 or x = 4, y = 2, i.e., the coordinates of intersection points are (1, - 1), (4, 2)
The area of the shaded part in the figure is ∫
Two
−1
(x+2−x2)dx＝(x2
2+2x−x3
3)
Two
−1
＝9
2．
Therefore, a

The area of the plane figure surrounded by the parabola y = 1 + x ^ 2, x = 0, x = 1 and y = 0 is calculated, and the volume of the body of revolution is obtained by rotating the figure around the X axis Thank you very much. 100 points reward, advanced mathematics first stage topic, difficulty is not big 28∏/15 There are two people who answer this question. The others are different. This question is not difficult. Which one is right? Who else knows,

The integral of area s to the integrand (x ^ 2 + 1) from 0 to 1,
That is, the difference between (1 / 3x ^ 3 + x) at 1 and 0, that is, s = 4 / 3
Volume is also integral
In this case, the integrand is p (x ^ 2 + 1) ^ 2 from 0 to 1 for X
I don't know how to input PI, which is represented by P
The results were 28P / 15
I don't know if you know
I made a mistake yesterday. I make up for it today
This result is absolutely correct

Find the area of the plane figure surrounded by the straight line y = 0, x = 0, x = 1 and the curve y = x cubic + 1 and the rotation of the figure around the X axis

S=∫(0→1) (x³+1) dx=[(x^4)/4+x]|(0→1)=5/4
V=∫(0→1) π(x³+1)² dx=π∫(0→1) (x^6+2x³+1) dx=π[(x^7)/7+(x^4)/2+x]|(0→1)=23π/14