- #1
Sociomath
- 9
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Hi,
I just need these solutions checked.
Thank you in advance!
Consider the region bounded by the following curves ##y=x-3, y=5-x, \text{and}\ y=3##:
1.) set up an integral expression that would give the area of the region of y as a function of x:
##y = x-3 = 5-x##
##x + x - 3 - 5 = 0##
##2x-8=0##
##x = 4##
##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##
2.) set up an integral expression that would give the area of the region of x as a function of y:
##\displaystyle \int_0^5 |\left(5-y\right)|##
3.) compute the area of the region by evaluating one of the expressions in (1.) or (2.):
##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##
##\left(x-8\right)+\ \text{C}##
##\displaystyle = \lim_{x \to 0^{+}} ((x-8)x)=0##
##\displaystyle = \lim_{x \to 3^{-}} ((x-8)x)=-15##
##\displaystyle |-15-0| = 15##
4.) set up an integral expression that would give the volume of the solid created by rotating the region about y = 1, using the disk/washer method:
##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
5.) find the volume of the solid by evaluating the integral/integrals in (4.):
##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
##=\ -\dfrac{29}{6}\pi##
##\approx 15\ \text{cubic units}##
6.) set up an integral expression that would give the volume of the solid created by rotating the region about y = -2, using the disk/washer method:
##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
7.) compute the volume of the solid by evaluating the integral or integrals in (6.):
##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
##=\ -\dfrac{27}{2}\pi##
##\approx 42\ \text{cubic units}##
I just need these solutions checked.
Thank you in advance!
Consider the region bounded by the following curves ##y=x-3, y=5-x, \text{and}\ y=3##:
1.) set up an integral expression that would give the area of the region of y as a function of x:
##y = x-3 = 5-x##
##x + x - 3 - 5 = 0##
##2x-8=0##
##x = 4##
##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##
2.) set up an integral expression that would give the area of the region of x as a function of y:
##\displaystyle \int_0^5 |\left(5-y\right)|##
3.) compute the area of the region by evaluating one of the expressions in (1.) or (2.):
##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##
##\left(x-8\right)+\ \text{C}##
##\displaystyle = \lim_{x \to 0^{+}} ((x-8)x)=0##
##\displaystyle = \lim_{x \to 3^{-}} ((x-8)x)=-15##
##\displaystyle |-15-0| = 15##
4.) set up an integral expression that would give the volume of the solid created by rotating the region about y = 1, using the disk/washer method:
##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
5.) find the volume of the solid by evaluating the integral/integrals in (4.):
##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
##=\ -\dfrac{29}{6}\pi##
##\approx 15\ \text{cubic units}##
6.) set up an integral expression that would give the volume of the solid created by rotating the region about y = -2, using the disk/washer method:
##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
7.) compute the volume of the solid by evaluating the integral or integrals in (6.):
##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##
##=\ -\dfrac{27}{2}\pi##
##\approx 42\ \text{cubic units}##