Answer to Question #205275 in Calculus for shane

Question

1. The curve has an equation y = e^x. Compute the area bounded by the curve from x = 0 to x = 1.

2. The loop of the curve has an equation of y² = x(1 – x)². Find the area enclosed by the loop of the curve.

3. Given the area in the first quadrant bounded by y² = x, the line x = 4 and the x-axis. What is the volume generated when this area is revolved about the y-axis?

4. The region in the first quadrant which is bounded by the curve y² = 4x, and the lines x = 4 and y = 0, is revolved about the x-axis. Locate the centroid of the resulting solid of revolution.

5. The region in the first quadrant, which is bounded by the curve x² = 4y, the line x = 4, is revolved about the line x = 4. Locate the centroid of the resulting solid of revolution.

Answer 1

Solution.

1.

$S=\int_0^1 e^x dx=e^x|_0^1=e-1.$

2.

$S=2\int_0^1x\sqrt{1-x}dx$

Find $\int x\sqrt{1-x}dx,$ for this we do replacement

$1-x=u.$ Then $x=1-u, -dx=du,\newline x\sqrt{1-x}=(1-u)\sqrt{u}=\sqrt{u}-u^{\frac{3}{2}}.$

So, $\int x\sqrt{1-x}dx=-\int (\sqrt{u}-u^{\frac{3}{2}})du=\newline \frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}=\newline \frac{2}{5}(1-x)^{\frac{5}{2}}-\frac{2}{3}(1-x)^{\frac{3}{2}}.$

Therefore,

$S=2(\frac{2}{5}(1-x)^{\frac{5}{2}}-\frac{2}{3}(1-x)^{\frac{3}{2}})|_0^1=\newline \frac{8}{15}.$

3.

$V_y=2π\int_0^4 x\sqrt{x}dx\newline 2π\int_0^4 x^{\frac{3}{2}}dx\newline 2π\frac{2}{5}x^{\frac{5}{2}}|_0^4=\newline \frac{128}{5}π=25\frac{3}{5}π.$

4.

The volume of the body V formed by rotation around the axis Ox of the figure $a\leq x\leq b,y_1(x)\leq y \leq y_2(x),$ , where y1(x) and y2(x) are continuous non-negative functions, is equal to a certain interval from the difference of the square of the functions yi(x) by the variable x:

$V_x=π\int_a^b(y_2^2(x)-y_1^2(x))dx$

We will have

$V_x=π\int_0^4 (2\sqrt{x})^2dx=4π \frac{x^2}{2}|_0^4=\newline 32π.$

In this case, the centroid lies on the x-axis. The formula to get the centroid of the figure is:

$Vx^*=\int xdV.$

$32πx^*=\int_0^4 4πx^2dx$

$32x^*=\frac{4x^3}{3}|_0^4$

From here, centroid $x^*=\frac{8}{3}.$

5.

The volume of the body V formed by rotation around the line x=m of the figure $a\leq y\leq b,0\leq x\leq x(y)$ , where x(y) is a unique continuous function equal to the definite integral calculated by the formula

$V=π\int_a^b(a^2-x^2(y))dy$

$V_{x=4}=π\int_0^4(-(2\sqrt{y})^2+4^3)dy=π(-2y^{2}+16y)|_0^4=32π.$

In this case, the centroid lies on the y-axis. The formula to get the centroid of the figure is$Vy^*=\int ydV.$

$Vy^*=\int_0^4yπ2\sqrt{y}dy=\frac{128}{5}π.$

From here $y^*=\frac{4}{5}.$