Hello everyone, I am new to this forum. That being said, I am sure I'm about to ask for way too much. I have four calculus problems that I have completed, but I am not sure that I got the right answers. If anyway possible, I would be very appreciative if someone looked at my answers, and told me if they were correct, and if not, showed me how to do the problem correctly.

Two problems have to do with area of a 2-d region, and two problems have to do with volume of a rotated area.

For Problems 1 & 2:Write the integral, or integrals, for the area, and evaluate your integral(s).

Problem 1.R is the region bounded by y = $\displaystyle x^2$ +4, y = -2x, x = 0, and x = 1.

$\displaystyle R_{1}$ is the region bounded by x = 1 and x = $\displaystyle \frac{-1}{2}$y from - 2 to 0 on the y-axis.My Solution

A($\displaystyle R_{1}$) = $\displaystyle \int^0_{-2}$(1 + $\displaystyle \frac{1}{2}y$) dy = [y + $\displaystyle \frac{y^2}{4}$$\displaystyle ]^0_{-2}$ = 0 + 1 = 1

$\displaystyle R_{2}$ is the region bounded by x = 1 and x = 0 from 0 to 4 on the y-axis.

A($\displaystyle R_{2}$) = $\displaystyle \int^4_0$(1) dy = [y$\displaystyle ]^4_0$ = 4 - 0 = 4

$\displaystyle R_{3}$ is the region bounded by x = 1 and x = $\displaystyle \sqrt{y - 4}$ from 4 to 5 on the y-axis.

A($\displaystyle R_{3}$) = $\displaystyle \int^5_4$(1 - $\displaystyle \sqrt{y - 4}$) dy = [y - $\displaystyle \frac{2(y - 4)^{\frac{3}{2}}}{3}$$\displaystyle ]^5_4$ = (5 - $\displaystyle \frac{2}{3}$) - (4) = $\displaystyle \frac{1}{3}$

The area of R is A($\displaystyle R_{1}$) + A($\displaystyle R_{2}$) + A($\displaystyle R_{3}$)

1 + 4 + $\displaystyle \frac{1}{3}$ = $\displaystyle \frac{16}{3}$

Problem 2.R is the region bounded by $\displaystyle y^2$ = x + 1, and the line x + y - 5 = 0.

$\displaystyle R_{1}$ is the region bounded by y = $\displaystyle \sqrt{x + 1}$ and y = 0 from -1 to 3 on the x-axis.My Solution

A($\displaystyle R_{1}$) = $\displaystyle \int^3_{-1}$$\displaystyle \sqrt{x + 1}$ dx = [$\displaystyle \frac{2(x + 1)^{\frac{3}{2}}}{3}$$\displaystyle ]^3_{-1}$ = $\displaystyle \frac{16}{3}$ - 0 = $\displaystyle \frac{16}{3}$

$\displaystyle R_{2}$ is the region bounded by y = -$\displaystyle \sqrt{x + 1}$ and y = 0 from -1 to 3 on the x-axis.

A($\displaystyle R_{2}$) = $\displaystyle \int^3_{-1}$$\displaystyle \sqrt{x + 1}$ dx = [$\displaystyle \frac{2(x + 1)^{\frac{3}{2}}}{3}$$\displaystyle ]^3_{-1}$ = $\displaystyle \frac{16}{3}$ - 0 = $\displaystyle \frac{16}{3}$

$\displaystyle R_{3}$ is the region bounded by x + y - 5 = 0 and y = 0 from 3 to 5 on the x-axis.

A($\displaystyle R_{3}$) = $\displaystyle \int^5_3$(-x + 5) dx = [$\displaystyle \frac{-x^2}{2}$ + 5x$\displaystyle ]^5_3$ = ($\displaystyle \frac{-25}{2}$ + 25) - ($\displaystyle \frac{-9}{2}$ + 15) = 2

$\displaystyle R_{4}$ is the region bounded by y = 0 and y = -$\displaystyle \sqrt{x + 1}$ from 3 to 8 on the x-axis.

A($\displaystyle R_{4}$) = $\displaystyle \int^8_3$$\displaystyle \sqrt{x + 1}$ dx = [$\displaystyle \frac{2(x + 1)^{\frac{3}{2}}}{3}$$\displaystyle ]^8_3$ = 18 - $\displaystyle \frac{16}{3}$ = $\displaystyle \frac{38}{3}$

The area of R is A($\displaystyle R_{1}$) + A($\displaystyle R_{2}$) + A($\displaystyle R_{3}$) + A($\displaystyle R_{4}$)

$\displaystyle \frac{16}{3}$ + $\displaystyle \frac{16}{3}$ + 2 + $\displaystyle \frac{38}{3}$ = $\displaystyle \frac{76}{3}$

For Problems 3 & 4:R is the region bounded by y =$\displaystyle x^2$ + 2, y = x, x = 0, and x = 2. Use the indicated method to write an integral, or integrals, for the volume of the solid of revolution if R is revolved in the indicated way. Then calculate the volume using either method (a), or method (b).

*** I think I got these wrong, so any help whatsoever would be much appreciated ***

Problem 3.Revolve about the x-axis.a.Disk Methodb.Shell Method

a. Disk MethodMy Solution

V = $\displaystyle \pi$$\displaystyle \int^2_0$($\displaystyle (x^2 + 2)^2$ - $\displaystyle x^2$) dx

b. Shell Method (Solved with Shell Method)

V = 2$\displaystyle \pi$$\displaystyle \int^2_0$(2$\displaystyle x^2$ + 4) dx = 2$\displaystyle \pi$[$\displaystyle \frac{2x^3}{3}$ + 4x$\displaystyle ]^2_0$ = 2$\displaystyle \pi$($\displaystyle \frac{16}{3}$ + 8) = $\displaystyle \frac{80\pi}{3}$

Problem 4.Revolve about the y-axis.a.Disk Methodb.Shell Method

a. Disk MethodMy Solution

V = $\displaystyle \pi$$\displaystyle \int^2_0$$\displaystyle (y)^2$ dy + $\displaystyle \pi$$\displaystyle \int^6_2$$\displaystyle (2 - \sqrt{2 - y})^2$ dy = $\displaystyle \pi$$\displaystyle \int^2_0$$\displaystyle y^2$ dy + $\displaystyle \pi$$\displaystyle \int^6_2$$\displaystyle (2 - \sqrt{2 - y})^2$ dy

b. Shell Method (Solved with Shell Method)

V = 2$\displaystyle \pi$$\displaystyle \int^2_0$2y dy + 2$\displaystyle \pi$$\displaystyle \int^6_2$8 dy = 2$\displaystyle \pi$[$\displaystyle y^2$$\displaystyle ]^2_0$ + 2$\displaystyle \pi$[8y$\displaystyle ]^6_2 $= 2$\displaystyle \pi$(4) + 2$\displaystyle \pi$(48 - 16) = 8$\displaystyle \pi$ + 64$\displaystyle \pi$ = 72$\displaystyle \pi$

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Once again, any help would be greatly appreciate. Thank you very much in advance.