Geometric application of assignment 27 definite integral

1.
(1) solving two curve equations to get the intersection of $ (0,0) $, $ (1,e) $, so the surrounding area
\[
A= \int_0 ^1 (x e-xe^x) dx
= \frac{e}{2} x^2 \bigg|_0^1-x e^x \bigg|_{0}^1 +\int_0^1 e^x DX =\frac e 2-1.
\]
(Note: When $0<x<1$ $e ^x < e$)

(2) Easy to solve the curve with the $x $ axis has three intersections, respectively, is $-1,0,2$, the area of the circumference
\[
A= \int_{-1}^0 (x^3-x^2-2x) dx + \int_0^2 (-x^3+x^2 +2x) dx = \frac 5{12} + \frac 8 3 = \frac{37}{12}.
\]

(3) It can be seen that this is a heart-shaped line, just the sharp angle of the heart facing up, it is about $y $ axis symmetry, so
\[
A= 2 \int_{\frac \pi 2}^{\frac32 \pi} \frac12 a^2 (1+\sin \theta) ^2 D\theta = \frac32 \pi a^2.
\]

2. The enclosed area is
\[
A (a) =\int_{a}^{a+1} x^2 DX =\frac13 (a+1) ^3-\frac13 a^3=a^2+a+\frac13,
\]
by $A ' (A) =0$ $a =-\frac 12$, when $a =-\frac12$
\[
\min A = \frac1{12}.
\]

3.

(1) to $x $ for the integral variable, the volume element can be
\[
DV = \pi (e^x-1) ^2 DX,
\]
So
\[
V =\int_0^{\ln 3} \pi (e^x-1) ^2 dx=\pi \ln 3.
\]

(2) The $y $ is the integral variable, the volume element can be
\[
DV = \pi (4-y) dy
\]
So
\[
v=\int_0^{4} \pi (4-y) dy =8 \pi.
\]

4. According to the question set, to $x $ for the integral variable, the observation image we just need to calculate $x $ from $0$ to $\pi a$, and then multiplied by two
\[
dv= [\PI (2a) ^2-\pi y^2 (x)] dx
\]
The
\[
v= 2\int_0^{\pi A} [\pi (2a) ^2-\pi y^2 (x)] DX.
\]
Using parametric equations to
\[
v= 2\int_0^{\pi} [\PI (2a) ^2-a^2 (1-\cos t) ^2]d (A (T-\sin t))
= 2\pi a^3 \int_0^{\pi} [(1-\cos t) ^2] (1-\cos t) DT =7\pi^2 a^3.
\]
or $y $ as the integral variable, the
\[
DV = 2 \pi (2a-y) (\pi-x) dy
\]
That
\[
v= \INT_0^{2A} 2 \pi (2a-y) (\pi-x) dy.
\]
Using parametric equations to
\[
v= 2 \int_0^{\pi} 2\pi [2a-a (1-\cos t)][\pi a-a (T-\sin t)] \cdot a \sin t DT
= 7\pi^2 a^3.
\]

5. The question was set
\[
Y ' =
\begin{cases}
-\frac 1x, & 1/e \leq x \leq 1,
\\
\frac 1 x, & 1<x\leq E.
\end{cases}
\]
The
\[
S = \int_{1/e}^1 \sqrt{1+ \frac 1{x^2}} DX +\int_1^{e} \sqrt{1+ \frac 1{x^2}} dx= \int_{1/e}^e \frac{\sqrt{1+x^2}}{x} dx.
\]
Because
\[
\int \frac{\sqrt{1+x^2}}{x} dx = \int \frac{1+x^2}{x \sqrt{1+x^2}} DX
=\int \frac{1}{x\sqrt{1+x^2}} DX + \sqrt{1+x^2},
\]
Make $t =\frac 1x$, then
\[
\int \frac{1}{x \sqrt{1+x^2}} dx =-\int \frac{1}{\sqrt{1+t^2}} dt =-\ln | t+ \sqrt{1+t^2} | =-\ln| X+\SQRT{1+X^2} | +\ln |x|.
\]
So
\[
\begin{aligned}
\int_{1/e}^e \frac{\sqrt{1+x^2}}{x} dx= \left (
\ln x +\sqrt{1+x^2}-\ln (x+\sqrt{1+x^2})
\right) \bigg|_{1/e}^{e}
\ =1+\sqrt{1+e^2}-\frac{\sqrt{1+e^2}}{e}-\ln (1+\sqrt{1+e^2})
+\ln (e+\sqrt{1+e^2}).
\end{aligned}
\]

6. The full length of the double-button line can be expressed as four times times the length of the first quadrant curve, so the parametric equation of the arc length can be obtained
\[
S = 4 \int_0^{\frac \pi 4} \sqrt{r^2 + (R ') ^2} D\theta
= 4 \sqrt 2 a \int_0^{\frac \pi 4} \sqrt{\cos 2\theta + \frac{\sin^2 2\theta}{\cos 2\theta}} D\theta
= 4 \sqrt 2 a \int_0^{\frac \pi 4} \frac{1}{\sqrt{\cos 2\theta}}d\theta
\]
and
\[
\int_0^{\frac \pi 4} \frac{1}{\sqrt{\cos 2\theta}}d\theta = \int_0^{\frac \pi 4} \frac{1}{\sqrt{\cos^2 \theta-\sin^2 \the Ta}}d\theta
= \int_0^{\frac \pi 4} \frac{\sec \theta}{\sqrt{1-\tan^2 \theta}}d\theta
\]
Make $x =\tan \theta$, then
\[
\int_0^{\frac \pi 4} \frac{\sec \theta}{\sqrt{1-\tan^2 \theta}}d\theta
= \int_0^1 \frac{\sqrt{1+x^2}}{\sqrt{1-x^2}} \cdot \frac{1}{1+x^2} dx =\int_0^1 \frac{1}{\sqrt{1-x^4}} dx.
\]
(How to see it ...) From the $x $ of the integral to do triangular transformation backward, the conclusion of the structure of the hint either do $x =\tan \theta$, or do $x =\sin \theta$, but if do $\sin $ transformation, the range and our polar coordinates calculated the integral does not correspond)

Geometric application of assignment 27 definite integral