# Answer in Calculus for shane #205275

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 y^{2} = x(1 – x)^{2}. Find the area enclosed by the loop of the curve.

3. Given the area in the first quadrant bounded by y^{2} = 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 y^{2} = 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 x^{2} = 4y, the line x = 4, is revolved about the line x = 4. Locate the centroid of the resulting solid of revolution.

Solution.

1.

“S=int_0^1 e^x dx=e^x|_0^1=e-1.”

2.

“S=2int_0^1xsqrt{1-x}dx”

Find “int xsqrt{1-x}dx,” for this we do replacement

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

So, “int xsqrt{1-x}dx=-int (sqrt{u}-u^{frac{3}{2}})du=newline nfrac{2}{5}u^{frac{5}{2}}-frac{2}{3}u^{frac{3}{2}}=newlinenfrac{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=2u03c0int_0^4 xsqrt{x}dxnewlinen2u03c0int_0^4 x^{frac{3}{2}}dxnewlinen2u03c0frac{2}{5}x^{frac{5}{2}}|_0^4=newlinenfrac{128}{5}u03c0=25frac{3}{5}u03c0.”

4.

The volume of the body V formed by rotation around the axis Ox of the figure “aleq xleq 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=u03c0int_a^b(y_2^2(x)-y_1^2(x))dx”

We will have

“V_x=u03c0int_0^4 (2sqrt{x})^2dx=4u03c0 frac{x^2}{2}|_0^4=newlinen32u03c0.”

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

“Vx^*=int xdV.”

“32u03c0x^*=int_0^4 4u03c0x^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 “aleq yleq b,0leq xleq x(y)” , where x(y) is a unique continuous function equal to the definite integral calculated by the formula

“V=u03c0int_a^b(a^2-x^2(y))dy”

“V_{x=4}=u03c0int_0^4(-(2sqrt{y})^2+4^3)dy=u03c0(-2y^{2}+16y)|_0^4=32u03c0.”

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_0^4yu03c02sqrt{y}dy=frac{128}{5}u03c0.”

From here “y^*=frac{4}{5}.”